Продолжение предметного доказательства ВТФ при степени
Возможны два метода доказательства, кроме метода Ферма (бесконечного спуска):
- метод аналогичный методу исследования ВТФ при степени три,
- метод исследования уравненя
, без метода Ферма.
Для доказательства ВТФ используем уравнение
Разложение уравнения
Вводом явной чётности переменных, имеем:
Для доказательства ВТФ необхоимо исследовать первые два уравнения.
Исследование уравнения:![$\[\left( {{\text{2x}}_{\text{1}} } \right)^m = \left( {{\text{2z}}_{\text{1}} + {\text{1}}} \right)^4 - \left( {{\text{2y}}_{\text{1}} + {\text{1}}} \right)^4 \]$](https://dxdy-04.korotkov.co.uk/f/b/8/e/b8e2efb6e57691da0b1ebb9973ec855682.png "$\[\left( {{\text{2x}}_{\text{1}} } \right)^m = \left( {{\text{2z}}_{\text{1}} + {\text{1}}} \right)^4 - \left( {{\text{2y}}_{\text{1}} + {\text{1}}} \right)^4 \]$")
Исследование возможно при использовании частных свойств уравнения.
Применение методов доказательства требуют введения понятий, определяющих некоторые свойства полиномиальных диофантова уравнений:- уравнение неопределённое при значении
- уравнение определённое при значении
- уравнение переопределённое при значении
-
значение степени
-
число неизвестных
При степени
уравнение переопределённое, ибо значение степени
больше числа неизвестных
. Так как число неизвестных два, уравнение определено при двух линейных сомножителях. Решение переопределённого уравнения Ферма возможно, если полученные из двух линейных сомножителей значения переменных в третьем сомножителе дают отвечающее оночлену число степени m. Из сказанного следует, что при одночлене с одним неизвестным и двух неизвестных
показатели степени сомножителей неоднородного уравнения порознь равны m, и можем писать:
![$$\[x_1^m = {\text{V}}_{\text{1}}^{\text{m}} {\text{V}}_{\text{2}}^{\text{m}} {\text{V}}_{\text{3}}^{\text{m}} ,{\text{(V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{,V}}_{\text{3}} ) = {\text{d}} = 1.\]$$](https://dxdy-04.korotkov.co.uk/f/f/3/6/f36860b87ce6c91a056a649ebf067f4c82.png "$$\[x_1^m = {\text{V}}_{\text{1}}^{\text{m}} {\text{V}}_{\text{2}}^{\text{m}} {\text{V}}_{\text{3}}^{\text{m}} ,{\text{(V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{,V}}_{\text{3}} ) = {\text{d}} = 1.\]$$")
Приведением к неоднородному виду, при использовании частных свойств, имеем:
![$$\[{\text{x}}_{\text{1}}^m = \frac{{\left( {2{\text{z}}_{\text{1}} + 1} \right)^4 - \left( {2{\text{y}}_{\text{1}} + 1} \right)^4 }}{{{\text{2}}^m }} = \frac{{\left( {\frac{{2{\text{z}}_{\text{1}} + 1}}
{2}} \right)^4 - \left( {\frac{{2y_{\text{1}} + 1}}{2}} \right)^4 }}
{{{\text{2}}^{m - 4} }} = \]$$](https://dxdy-03.korotkov.co.uk/f/a/6/4/a644833d0ca5daf936a9272d94b1269982.png "$$\[{\text{x}}_{\text{1}}^m = \frac{{\left( {2{\text{z}}_{\text{1}} + 1} \right)^4 - \left( {2{\text{y}}_{\text{1}} + 1} \right)^4 }}{{{\text{2}}^m }} = \frac{{\left( {\frac{{2{\text{z}}_{\text{1}} + 1}}
{2}} \right)^4 - \left( {\frac{{2y_{\text{1}} + 1}}{2}} \right)^4 }}
{{{\text{2}}^{m - 4} }} = \]$$")
![$$\[ = \frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}{2}} \right)}}{{2^{m - 4} }} = {\text{V}}_{\text{1}}^{\text{m}} {\text{V}}_{\text{2}}^{\text{m}} {\text{V}}_{\text{3}}^{\text{m}} ,\]$$](https://dxdy-01.korotkov.co.uk/f/0/c/8/0c8ed8d2d29e6b1ceeeb91db26d8548282.png "$$\[ = \frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}{2}} \right)}}{{2^{m - 4} }} = {\text{V}}_{\text{1}}^{\text{m}} {\text{V}}_{\text{2}}^{\text{m}} {\text{V}}_{\text{3}}^{\text{m}} ,\]$$")
На основе 1. и 2. варианта, решений нет, на основе 3. варианта, нет при степени
,
дробь. Слеовательно, по первому методу доказательства уравнение не имеет требуемых решений. Случаи
исследуемы отдельно.
Исследование 3-го варианта решений при степени 
(второй метод доказательства)
Из первых двух уравнений 3-го варианта получаем значения переменных:
![$$\[{\text{ y}}_{\text{1}} = \frac{{{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} - {\text{1}}}}
{{\text{2}}}{\text{, y}} = {\text{2y}}_{\text{1}} + {\text{1}} = {\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} {\text{, z}}_{\text{1}} = \frac{{{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} - {\text{1}}}}
{{\text{2}}}{\text{,z}} = {\text{2z}}_{\text{1}} + {\text{1}} = {\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} {\text{. }}\]$$](https://dxdy-04.korotkov.co.uk/f/3/1/4/3148b857453c0992ea07ab521559beca82.png "$$\[{\text{ y}}_{\text{1}} = \frac{{{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} - {\text{1}}}}
{{\text{2}}}{\text{, y}} = {\text{2y}}_{\text{1}} + {\text{1}} = {\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} {\text{, z}}_{\text{1}} = \frac{{{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} - {\text{1}}}}
{{\text{2}}}{\text{,z}} = {\text{2z}}_{\text{1}} + {\text{1}} = {\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} {\text{. }}\]$$")
Подставляя значения переменных в 3. уравнение 3. варианта решений, имеем:
![$$\[{\text{V}}_{\text{3}}^{\text{2}} = \left[ {{\text{4}}\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} } \right) + 4\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} } \right) + 2} \right] = {\text{z}}^{\text{2}} + {\text{y}}^{\text{2}} = \left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 = 2\left( {\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right),\]$$](https://dxdy-02.korotkov.co.uk/f/d/f/e/dfe5a867bccd36e756f68d991bc5617f82.png "$$\[{\text{V}}_{\text{3}}^{\text{2}} = \left[ {{\text{4}}\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} } \right) + 4\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} } \right) + 2} \right] = {\text{z}}^{\text{2}} + {\text{y}}^{\text{2}} = \left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 = 2\left( {\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right),\]$$")
![$\[{\text{V}}_{\text{2}}^{\text{2}} > \left( {{\text{V}}_{\text{1}}^{\text{2}} + 1} \right),\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$](https://dxdy-04.korotkov.co.uk/f/3/7/2/372d17b7cec58a1a5e5016033134fdef82.png "$\[{\text{V}}_{\text{2}}^{\text{2}} > \left( {{\text{V}}_{\text{1}}^{\text{2}} + 1} \right),\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$")
числа различной чётности,
![$\[
{\text{V}}_{\text{3}}^{\text{2}} - \]$](https://dxdy-04.korotkov.co.uk/f/f/e/6/fe6be511b962395410ab4343afb36d6b82.png "$\[
{\text{V}}_{\text{3}}^{\text{2}} - \]$")
дробь.
Правая сторона уравнения чётная, значит
![$\[{\text{V}}_{\text{3}}\]$](https://dxdy-02.korotkov.co.uk/f/d/7/c/d7c42e46e2b0320052ac2c6b9d508e5b82.png "$\[{\text{V}}_{\text{3}}\]$")
тоже чётное число:
![$$\[{\text{V}}_{\text{3}}^{\text{2}} = \left( {{\text{2V}}_{\text{3}}^{\text{'}} } \right)^2 = 2\left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right),{\text{ V}}_{\text{3}}^{{\text{'2}}} = \frac{{{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} }}{2}.\]$$](https://dxdy-02.korotkov.co.uk/f/1/7/0/1703f89de94c907896fc596aafb2f17582.png "$$\[{\text{V}}_{\text{3}}^{\text{2}} = \left( {{\text{2V}}_{\text{3}}^{\text{'}} } \right)^2 = 2\left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right),{\text{ V}}_{\text{3}}^{{\text{'2}}} = \frac{{{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} }}{2}.\]$$")
Чётность
![$\[\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$](https://dxdy-01.korotkov.co.uk/f/4/9/1/491e8b413281f2f68de03780e4a61d0282.png "$\[\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$")
различная, и
![$\[{\text{V}}_{\text{3}}^{\text{'}} \]$](https://dxdy-03.korotkov.co.uk/f/a/9/d/a9daa0321e8d4f449688b67e97e160da82.png "$\[{\text{V}}_{\text{3}}^{\text{'}} \]$")
не натуральное число. Это исключает разрешимость уравнения согласно требованиям при степени
следовательно, и при значении степени
Подстановкой в уравнение, проверим равенство:
![$$\[ = \left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 - \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 } \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 } \right] = \]$$](https://dxdy-04.korotkov.co.uk/f/b/7/4/b74b8b8226b3c3a56da278d68817978482.png "$$\[ = \left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 - \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 } \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 } \right] = \]$$")
![$$\[ = \left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right) - \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)} \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right) + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)} \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 } \right] = \]$$](https://dxdy-04.korotkov.co.uk/f/3/f/9/3f9a42bf7813dedbd87272194a76e06282.png "$$\[ = \left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right) - \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)} \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right) + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)} \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 } \right] = \]$$")
![$$\[
= 2{\text{V}}_{\text{1}}^{\text{2}} {\text{2V}}_{\text{2}}^{\text{2}} {\text{2}}\left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right) = \left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^2.\]$$](https://dxdy-01.korotkov.co.uk/f/c/7/7/c77a79a115e2360bd9772dccdb57c66b82.png "$$\[
= 2{\text{V}}_{\text{1}}^{\text{2}} {\text{2V}}_{\text{2}}^{\text{2}} {\text{2}}\left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right) = \left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^2.\]$$")
Исследование 3-го варианта решений при степени 
(для доказательства ВТФ не требуется)
Из первых двух уравнений 3-го варианта получаем значения переменных:
![$$\[{\text{z}}_{\text{1}} = \frac{{{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} - {\text{1}}}}
{{\text{2}}}{\text{ }}{\text{, z}} = {\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} {\text{, y}}_{\text{1}} = \frac{{{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} - 1}}
{2}{\text{ }}{\text{, y}} = {\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}}.\]$$](https://dxdy-01.korotkov.co.uk/f/4/2/e/42e7c049194176577b51e94678b5c06882.png "$$\[{\text{z}}_{\text{1}} = \frac{{{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} - {\text{1}}}}
{{\text{2}}}{\text{ }}{\text{, z}} = {\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} {\text{, y}}_{\text{1}} = \frac{{{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} - 1}}
{2}{\text{ }}{\text{, y}} = {\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}}.\]$$")
Поставляя значения переменных в 3. уравнение 3. варианта решений, имеем:
![$$\[{\text{2V}}_{\text{3}}^{\text{3}} = \left[ {4\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} } \right) + 4\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} } \right) + 1} \right] = {\text{z}}^{\text{2}} + {\text{y}}^{\text{2}} = \left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 = 2\left( {{\text{V}}_{\text{2}}^{\text{6}} + {\text{V}}_{\text{1}}^{\text{6}} } \right){\text{,}}\]$$](https://dxdy-04.korotkov.co.uk/f/b/7/9/b793a08dd5f7007f51fdf90104fe152182.png "$$\[{\text{2V}}_{\text{3}}^{\text{3}} = \left[ {4\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} } \right) + 4\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} } \right) + 1} \right] = {\text{z}}^{\text{2}} + {\text{y}}^{\text{2}} = \left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 = 2\left( {{\text{V}}_{\text{2}}^{\text{6}} + {\text{V}}_{\text{1}}^{\text{6}} } \right){\text{,}}\]$$")
![$\[{\text{V}}_{\text{3}}^{\text{3}} = \left( {V_2^2 } \right)^3 + \left( {V_1^2 } \right)^3 ,{\text{V}}_{\text{2}}^{\text{3}} > {\text{V}}_{\text{1}}^{\text{3}} + {\text{1}}{\text{,}}\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$](https://dxdy-02.korotkov.co.uk/f/9/1/d/91dff3d8c6640753b6fe80fab1ad86d782.png "$\[{\text{V}}_{\text{3}}^{\text{3}} = \left( {V_2^2 } \right)^3 + \left( {V_1^2 } \right)^3 ,{\text{V}}_{\text{2}}^{\text{3}} > {\text{V}}_{\text{1}}^{\text{3}} + {\text{1}}{\text{,}}\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$")
различной чётности,
![$\[{\text{V}}_{\text{3}}^{\text{3}} - \]$](https://dxdy-03.korotkov.co.uk/f/e/b/6/eb64e0d04b20bdb8990e05bbe624baa482.png "$\[{\text{V}}_{\text{3}}^{\text{3}} - \]$")
дробь.
Исходя из полученного уравнения,
![$\[{\text{V}}_{\text{3}} \]$](https://dxdy-03.korotkov.co.uk/f/e/a/7/ea794c62a85d2696782bcf3a9741bfa882.png "$\[{\text{V}}_{\text{3}} \]$")
не натуральное число. Это исключает разрешимость уравнения согласно требованиям при степени
Подстановкой в уравнение, проверим равенство:
![$$\[\left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^3 = \left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^4 - \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^4 = \]$$](https://dxdy-01.korotkov.co.uk/f/8/9/e/89e798102bcc8c01ed8b714970b0c92082.png "$$\[\left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^3 = \left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^4 - \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^4 = \]$$")
![$$\[ = \left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 - \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 } \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 } \right] = \]$$](https://dxdy-01.korotkov.co.uk/f/4/9/5/495aee0b55034fb7fd17f696657468ba82.png "$$\[ = \left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 - \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 } \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 } \right] = \]$$")
![$$\[= \left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right) - \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)} \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right) + \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)} \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 } \right] =\]$$](https://dxdy-01.korotkov.co.uk/f/4/8/6/4864077386a3612a2c4cc41f8af612b282.png "$$\[= \left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right) - \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)} \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right) + \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)} \right]\left[ {\left( {{\text{V}}_{\text{2}}^{\text{3}} + {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{3}} - {\text{V}}_{\text{1}}^{\text{3}} } \right)^2 } \right] =\]$$")
![$$\[ = 2{\text{V}}_{\text{1}}^{\text{3}} {\text{2V}}_{\text{2}}^{\text{3}} {\text{2}}\left( {{\text{V}}_{\text{2}}^{\text{6}} + {\text{V}}_{\text{1}}^{\text{6}} } \right) = \left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^3.\]$$](https://dxdy-04.korotkov.co.uk/f/b/8/3/b83a831edd64f867bef5126d6b4aeefb82.png "$$\[ = 2{\text{V}}_{\text{1}}^{\text{3}} {\text{2V}}_{\text{2}}^{\text{3}} {\text{2}}\left( {{\text{V}}_{\text{2}}^{\text{6}} + {\text{V}}_{\text{1}}^{\text{6}} } \right) = \left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^3.\]$$")
Уравнение не имеет неоднородных решений при степени
но имеет множество неоднородных решений при степени
Уравнение имеет также партикулярно однородные решений при
При
![$\[{\text{m}} = {\text{n}}\]$](https://dxdy-01.korotkov.co.uk/f/c/3/7/c3750840d1da2387098e2b10559fdef882.png "$\[{\text{m}} = {\text{n}}\]$")
таких решений нет, ибо уравнение Ферма имело бы решение при степени
Получение партикулярно однородных решений уравнения можно, исходя из неоднородных уравнений, имеющих решение, следующим образом:
![$\[{\text{x}}^{{\text{q }}} = z^{\text{n}} - y^{\text{n}} ,{\text{x}}^{\text{q}} {\text{d}} = z^{\text{n}} d - y^{\text{n}} d,x^q = z^{\text{n}} - {\text{y}}^{\text{n}} - \]$](https://dxdy-02.korotkov.co.uk/f/5/3/4/534ddec1a974fd8b1f36387b2b29786582.png "$\[{\text{x}}^{{\text{q }}} = z^{\text{n}} - y^{\text{n}} ,{\text{x}}^{\text{q}} {\text{d}} = z^{\text{n}} d - y^{\text{n}} d,x^q = z^{\text{n}} - {\text{y}}^{\text{n}} - \]$")
уравнение, имеющее решение,
После подстановки получаем:
![$$\[{\text{x}}^{\text{q}} {\text{x}}^{{\text{kq}}} ^{\text{n}} = {\text{z}}^n {\text{x}}^{{\text{kqn}}} - {\text{y}}^n {\text{x}}^{{\text{kqn}}} {\text{, }}x^{{\text{q ( kn}} + {\text{1)}}} = x^m = \left( {{\text{zx}}^{{\text{kq }}} } \right)^{\text{n}} - \left( {{\text{yx}}^{{\text{kq }}} } \right)^{\text{n}} ,k = 1,2,3, \cdot \cdot \cdot ,m = {\text{q ( kn}} + {\text{1)}}{\text{.}}\]$$](https://dxdy-04.korotkov.co.uk/f/7/3/9/73927da49c722f9e5a10ad358701fd9082.png "$$\[{\text{x}}^{\text{q}} {\text{x}}^{{\text{kq}}} ^{\text{n}} = {\text{z}}^n {\text{x}}^{{\text{kqn}}} - {\text{y}}^n {\text{x}}^{{\text{kqn}}} {\text{, }}x^{{\text{q ( kn}} + {\text{1)}}} = x^m = \left( {{\text{zx}}^{{\text{kq }}} } \right)^{\text{n}} - \left( {{\text{yx}}^{{\text{kq }}} } \right)^{\text{n}} ,k = 1,2,3, \cdot \cdot \cdot ,m = {\text{q ( kn}} + {\text{1)}}{\text{.}}\]$$")
Для доказательства ВТФ необходимо доказать, что второе уравнение также не имеет требуемых решений.
Исследование уравнения 
Запишем уравнение при степени
![$\[\left( {2z_1 } \right)^2 = \left( {{\text{2x}}_{\text{1}} + {\text{1}}} \right)^{\text{2}} + \left( {{\text{2y}}_{\text{1}} + {\text{1}}} \right)^{\text{2}}.\]$](https://dxdy-01.korotkov.co.uk/f/4/d/5/4d596e9002fcaa72564bcc0864e1b3d982.png "$\[\left( {2z_1 } \right)^2 = \left( {{\text{2x}}_{\text{1}} + {\text{1}}} \right)^{\text{2}} + \left( {{\text{2y}}_{\text{1}} + {\text{1}}} \right)^{\text{2}}.\]$")
Приведением уравнения к неоднородному виду, имеем:
![$$\[z_1^2 = \frac{{\left( {{\text{2x}}_{\text{1}} + {\text{1}}} \right)^{\text{2}} + \left( {{\text{2y}}_{\text{1}} + {\text{1}}} \right)^{\text{2}} }}{4} = x_1^2 + x_1 + y_1^2 + y_1 + \frac{1}
{2}.\]$$](https://dxdy-03.korotkov.co.uk/f/2/7/5/275225a63b5d2347114e23706521e34982.png "$$\[z_1^2 = \frac{{\left( {{\text{2x}}_{\text{1}} + {\text{1}}} \right)^{\text{2}} + \left( {{\text{2y}}_{\text{1}} + {\text{1}}} \right)^{\text{2}} }}{4} = x_1^2 + x_1 + y_1^2 + y_1 + \frac{1}
{2}.\]$$")
Дробное значение
исключает разрешимость уравнения при значениях
.
Из результатов исследования двух уравнений следует, что ВТФ при степени не имеет требуемых решений. 
определено двумя линейными сомножителями, генерирующими все возможные натуральные значения переменных 
в остаточном сводном сомножителе генерируют число степени m. Ибо при одночлене с одним неизвестным и 
с двумя неизвестными для переопределённого ![$\[x_1^m = {\text{V}}_{\text{1}}^{\text{m}} {\text{V}}_{\text{2}}^{\text{m}} {\text{V}}_{\text{3}}^{\text{m}} .
\]$](https://dxdy-03.korotkov.co.uk/f/2/0/1/2013fb1331f5f339f09a6cfbc403e42b82.png "$\[x_1^m = {\text{V}}_{\text{1}}^{\text{m}} {\text{V}}_{\text{2}}^{\text{m}} {\text{V}}_{\text{3}}^{\text{m}} .
\]$")
И совершенно иной вопрос, что равенство возможно при натуральном или ненатуральном значении ![$\[{\text{V}}_{\text{3}} .\]$](https://dxdy-03.korotkov.co.uk/f/6/0/1/6019dc9583017273fa53037bae70019782.png "$\[{\text{V}}_{\text{3}} .\]$")
определено тремя, или соответственно большим числом линейных сомножителей, генерирующими все возможные натуральные значения переменных 
с k неизвестными для переопределённого 
сомножителями, имеем:![$$\[x_1^m = {\text{V}}_{\text{1}} {\text{V}}_{\text{2}} \cdot \cdot \cdot {\text{V}}_{\text{k}} {\text{V}}_{{\text{k}} + {\text{1}}} ,{\text{(V}}_{\text{1}} {\text{,V}}_{\text{2}} , \cdot \cdot \cdot ,{\text{V}}_{\text{k}} {\text{,V}}_{{\text{k}} + {\text{1}}} ) = {\text{d}} = 1.\]$$](https://dxdy-03.korotkov.co.uk/f/6/c/9/6c96ea61f60e506b885aadf3532428f882.png "$$\[x_1^m = {\text{V}}_{\text{1}} {\text{V}}_{\text{2}} \cdot \cdot \cdot {\text{V}}_{\text{k}} {\text{V}}_{{\text{k}} + {\text{1}}} ,{\text{(V}}_{\text{1}} {\text{,V}}_{\text{2}} , \cdot \cdot \cdot ,{\text{V}}_{\text{k}} {\text{,V}}_{{\text{k}} + {\text{1}}} ) = {\text{d}} = 1.\]$$")
Ведь сложное число степени m разлагается на ![$\[x_1^m = {\text{V}}_{\text{1}} {\text{V}}_{\text{2}} \cdot \cdot \cdot {\text{V}}_{\text{k}} {\text{V}}_{{\text{k + 1}}} .\]$](https://dxdy-02.korotkov.co.uk/f/1/8/f/18fc2522dd1337671dd557659f8062d082.png "$\[x_1^m = {\text{V}}_{\text{1}} {\text{V}}_{\text{2}} \cdot \cdot \cdot {\text{V}}_{\text{k}} {\text{V}}_{{\text{k + 1}}} .\]$")
![$$\[\frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}{2}} \right)}}{{2^{m - 4} }} = {\text{V}}_{\text{1}}^{\text{m}} {\text{V}}_{\text{2}}^{\text{m}} {\text{V}}_{\text{3}}^{\text{m}} ,\]$$](https://dxdy-01.korotkov.co.uk/f/8/a/9/8a91bdcf2f7e0c67a80d7f70ef58c8b282.png "$$\[\frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}{2}} \right)}}{{2^{m - 4} }} = {\text{V}}_{\text{1}}^{\text{m}} {\text{V}}_{\text{2}}^{\text{m}} {\text{V}}_{\text{3}}^{\text{m}} ,\]$$")
вместо 
используете чтобы легче было запутаться?![$\[\left( {a^4 - b^4 } \right)\]$](https://dxdy-03.korotkov.co.uk/f/a/a/1/aa1316abd672051cc1bed3fced6c92a982.png "$\[\left( {a^4 - b^4 } \right)\]$")
на множители, как и дробного выражения ![$\left[ {\left( {\frac{{2z_1 + 1}}{2}} \right)^4 - \left( {\frac{{2y_1 + 1}}{2}} \right)^4 } \right],$](https://dxdy-02.korotkov.co.uk/f/1/0/4/1046359ed1136f03240e75d97c3f8f6282.png "$\left[ {\left( {\frac{{2z_1 + 1}}{2}} \right)^4 - \left( {\frac{{2y_1 + 1}}{2}} \right)^4 } \right],$")
однозначно определено правилами разложения на множители, исходя из биномиальной теоремы Ньютона, других разложений не существует! В уравнении для ФТФ, при степени ![$\[m = n = 4,\]$](https://dxdy-01.korotkov.co.uk/f/8/e/e/8ee98b17ede3f138105d7df95255f3e682.png "$\[m = n = 4,\]$")
имеем: ![$$\[x_1^m = \frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}
{2}} \right)}}{{2^{m - 4} }} = \]$$](https://dxdy-01.korotkov.co.uk/f/c/0/9/c09ed92c002eb8ce23a3deae22bfd0c282.png "$$\[x_1^m = \frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}
{2}} \right)}}{{2^{m - 4} }} = \]$$")
![$$\[ = \frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}
{2}} \right)}}{{2^{4 - 4} }} = \]$$](https://dxdy-01.korotkov.co.uk/f/8/2/b/82bb50fe7f2e14353b749e47b121a3bc82.png "$$\[ = \frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}
{2}} \right)}}{{2^{4 - 4} }} = \]$$")
![$$\[ = \frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}
{2}} \right)}}{1} = \cdot \cdot \cdot.\]$$](https://dxdy-02.korotkov.co.uk/f/9/4/4/944ffa7c984befcd566bd5733cd343f282.png "$$\[ = \frac{{\left( {{\text{z}}_{\text{1}} - {\text{y}}_{\text{1}} } \right)\left( {{\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + 1} \right)\left( {{\text{z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + {\text{z}}_{\text{1}} + {\text{y}}_{\text{1}} + \frac{1}
{2}} \right)}}{1} = \cdot \cdot \cdot.\]$$")
![$ \[m \ne n = 4,\]$](https://dxdy-04.korotkov.co.uk/f/3/a/e/3aee36b073e9f4393553dc8d7856e5e682.png "$ \[m \ne n = 4,\]$")
ибо знаменатель дроби не входит в разложение и он может быть исключительно множителем ![$\[V_1^3 ,V_2^3 ,V_3^3 .\]$](https://dxdy-02.korotkov.co.uk/f/d/e/6/de676b65240eb24147ef91471fc294ea82.png "$\[V_1^3 ,V_2^3 ,V_3^3 .\]$")
Поэтому при знаменателе ![$ \[2^{m - n} \ne 1\]$](https://dxdy-03.korotkov.co.uk/f/a/9/0/a902c9ba40b0f8dbb2286a37c6b4bcb782.png "$ \[2^{m - n} \ne 1\]$")
значения ![$\[
V_1^3 ,V_2^3 ,V_3^3 \]$](https://dxdy-01.korotkov.co.uk/f/0/b/4/0b439b0c1878891ff18bb42340ccf35282.png "$\[
V_1^3 ,V_2^3 ,V_3^3 \]$")
можно соответственно умножить на значение ![$\[2^{m - n} .\]$](https://dxdy-04.korotkov.co.uk/f/b/9/1/b91c365060d8da044642b862875863e982.png "$\[2^{m - n} .\]$")
Это Вы правитьно показали в первом уравнении предложенного варианта. Там же показали неправомерный перенос двойки в знаменатель, нарушая правила розложения на множители, условия биномиальной теоремы. Знаменатель![$\[2^{m - n} \]$](https://dxdy-03.korotkov.co.uk/f/e/0/9/e0901a776be3cfb50d6e80366223e13282.png "$\[2^{m - n} \]$")
при степени ![$\[{\text{1}} < {\text{m}} < 4\]$](https://dxdy-02.korotkov.co.uk/f/5/d/6/5d6bc8a117b8c1c68737164f1ef42f5e82.png "$\[{\text{1}} < {\text{m}} < 4\]$")
приводит к натуральному значению третьего сомножителя, не равному квадрату, или кубу натурального числа. 
?
![$$\left( {2x_1 } \right)^2 = 4\left( {z_1 - y_1 } \right)\left( {z_1 + y_1 + 1} \right)\left[ {2\left( {{\text{2z}}_{\text{1}}^{\text{2}} + 2{\text{y}}_{\text{1}}^{\text{2}} + 2{\text{z}}_{\text{1}} + 2{\text{y}}_{\text{1}} + 1} \right)} \right],$$](https://dxdy-04.korotkov.co.uk/f/3/1/7/31762ba85cd00df6b6098088cb75373182.png "$$\left( {2x_1 } \right)^2 = 4\left( {z_1 - y_1 } \right)\left( {z_1 + y_1 + 1} \right)\left[ {2\left( {{\text{2z}}_{\text{1}}^{\text{2}} + 2{\text{y}}_{\text{1}}^{\text{2}} + 2{\text{z}}_{\text{1}} + 2{\text{y}}_{\text{1}} + 1} \right)} \right],$$")
![$$x_1^2 = \left( {z_1 - y_1 } \right)\left( {z_1 + y_1 + 1} \right)\left[ {2\left( {{\text{2z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + 2{\text{z}}_{\text{1}} + 2{\text{y}}_{\text{1}} + 1} \right)} \right] = V_1^2 V_2^2 V_3^2 ,$$](https://dxdy-04.korotkov.co.uk/f/b/d/a/bdac69e993336d81024f73aee093042482.png "$$x_1^2 = \left( {z_1 - y_1 } \right)\left( {z_1 + y_1 + 1} \right)\left[ {2\left( {{\text{2z}}_{\text{1}}^{\text{2}} + {\text{y}}_{\text{1}}^{\text{2}} + 2{\text{z}}_{\text{1}} + 2{\text{y}}_{\text{1}} + 1} \right)} \right] = V_1^2 V_2^2 V_3^2 ,$$")
![$$\[x_1^2 = {\text{V}}_{\text{1}}^{\text{2}} {\text{V}}_{\text{2}}^{\text{2}} {\text{V}}_{\text{3}}^{\text{2}} ,{\text{(V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{,V}}_{\text{3}} ) = {\text{d}} = 1,\]$$](https://dxdy-03.korotkov.co.uk/f/6/4/d/64d6835a3ed92705e4b913b9d075709382.png "$$\[x_1^2 = {\text{V}}_{\text{1}}^{\text{2}} {\text{V}}_{\text{2}}^{\text{2}} {\text{V}}_{\text{3}}^{\text{2}} ,{\text{(V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{,V}}_{\text{3}} ) = {\text{d}} = 1,\]$$")
![$\[
{\text{V}}_{\text{2}}^{\text{2}} > \left( {{\text{V}}_{\text{1}}^{\text{2}} + 1} \right),\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$](https://dxdy-01.korotkov.co.uk/f/0/d/c/0dcb98d0221f280c209db6d4f89339a382.png "$\[
{\text{V}}_{\text{2}}^{\text{2}} > \left( {{\text{V}}_{\text{1}}^{\text{2}} + 1} \right),\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$")
числа различной чётности, ![$\[{\text{V}}_{\text{3}}^{\text{2}} - \] $](https://dxdy-02.korotkov.co.uk/f/5/c/a/5ca89bcbbfb9323bab8f2b6d8be01a6682.png "$\[{\text{V}}_{\text{3}}^{\text{2}} - \] $")
дробь. 
![$$\[
{\text{V}}_{\text{3}}^{\text{2}} = {\text{z}}^{\text{2}} + {\text{y}}^{\text{2}} = \left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 = 2\left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right).\]$$](https://dxdy-01.korotkov.co.uk/f/4/a/2/4a2f54dadfd3aaa88730957afc06f15082.png "$$\[
{\text{V}}_{\text{3}}^{\text{2}} = {\text{z}}^{\text{2}} + {\text{y}}^{\text{2}} = \left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^2 = 2\left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right).\]$$")
![$\[
{\text{V}}_{\text{3}} \]$](https://dxdy-04.korotkov.co.uk/f/b/9/1/b910da74f91915b0dc4e0e501f43a9f382.png "$\[
{\text{V}}_{\text{3}} \]$")
тоже чётное число:![$\[{\text{V}}_{\text{3}}^{{\text{'2}}}\]$](https://dxdy-03.korotkov.co.uk/f/6/2/f/62f98f54dfa51d9aef68196a3bcc4cce82.png "$\[{\text{V}}_{\text{3}}^{{\text{'2}}}\]$")
дробное число. Это исключает разрешимость уравнения согласно требованиям при значении степени ![$$\[\left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^2 = \left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^4 - \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^4 = \]$$](https://dxdy-02.korotkov.co.uk/f/9/f/6/9f6fd566b6ca78320b090d6a616aad0c82.png "$$\[\left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^2 = \left( {{\text{V}}_{\text{2}}^{\text{2}} + {\text{V}}_{\text{1}}^{\text{2}} } \right)^4 - \left( {{\text{V}}_{\text{2}}^{\text{2}} - {\text{V}}_{\text{1}}^{\text{2}} } \right)^4 = \]$$")
![$$\[
= 2{\text{V}}_{\text{1}}^{\text{2}} {\text{2V}}_{\text{2}}^{\text{2}} {\text{2}}\left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right) = \left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^2 .\]$$](https://dxdy-02.korotkov.co.uk/f/5/f/1/5f17fe08efa716b654beb307a8ba237f82.png "$$\[
= 2{\text{V}}_{\text{1}}^{\text{2}} {\text{2V}}_{\text{2}}^{\text{2}} {\text{2}}\left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right) = \left( {{\text{2V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} } \right)^2 .\]$$")
Введением явной чётности переменных, имеем:![$$\left( {2x_1 } \right)^4 = 4\left( {z_1 - y_1 } \right)\left( {z_1 + y_1 + 1} \right)\left[ {2\left( {{\text{2z}}_{\text{1}}^{\text{2}} + 2{\text{y}}_{\text{1}}^{\text{2}} + 2{\text{z}}_{\text{1}} + 2{\text{y}}_{\text{1}} + 1} \right)} \right],$$](https://dxdy-02.korotkov.co.uk/f/1/1/9/119318c2f6841967a8cae41e254c19d282.png "$$\left( {2x_1 } \right)^4 = 4\left( {z_1 - y_1 } \right)\left( {z_1 + y_1 + 1} \right)\left[ {2\left( {{\text{2z}}_{\text{1}}^{\text{2}} + 2{\text{y}}_{\text{1}}^{\text{2}} + 2{\text{z}}_{\text{1}} + 2{\text{y}}_{\text{1}} + 1} \right)} \right],$$")
![$$\[
x_1^4 = {\text{V}}_{\text{1}}^{\text{4}} {\text{V}}_{\text{2}}^{\text{4}} {\text{V}}_{\text{3}}^{\text{4}} ,{\text{(V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{,V}}_{\text{3}} ) = {\text{d}} = 1,x = 2x_1 = 2{\text{V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} ,\]$$](https://dxdy-03.korotkov.co.uk/f/2/b/8/2b83d8dfca8105d2e9b4c623ea3ecacd82.png "$$\[
x_1^4 = {\text{V}}_{\text{1}}^{\text{4}} {\text{V}}_{\text{2}}^{\text{4}} {\text{V}}_{\text{3}}^{\text{4}} ,{\text{(V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{,V}}_{\text{3}} ) = {\text{d}} = 1,x = 2x_1 = 2{\text{V}}_{\text{1}} {\text{V}}_{\text{2}} {\text{V}}_{\text{3}} ,\]$$")
![$\[{\text{V}}_{\text{2}}^{\text{4}} > \left( {{\text{V}}_{\text{1}}^{\text{4}} + 1} \right),\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$](https://dxdy-04.korotkov.co.uk/f/7/7/8/778070aef36f3b8a0d48dcdec3f05db382.png "$\[{\text{V}}_{\text{2}}^{\text{4}} > \left( {{\text{V}}_{\text{1}}^{\text{4}} + 1} \right),\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} - \]$")
числа различной чётности, ![$\[
{\text{V}}_{\text{3}}^{\text{4}} - \]$](https://dxdy-02.korotkov.co.uk/f/1/9/a/19a1234b9a019e3c1b7b5abc4da6c49582.png "$\[
{\text{V}}_{\text{3}}^{\text{4}} - \]$")
дробь. 
![$$\[{\text{V}}_{\text{3}}^{\text{4}} = {\text{z}}^{\text{2}} + {\text{y}}^{\text{2}} = \left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{4}} - {\text{V}}_{\text{1}}^{\text{4}} } \right)^2 = 2\left( {{\text{V}}_{\text{2}}^{\text{8}} + {\text{V}}_{\text{1}}^{\text{8}} } \right).\]$$](https://dxdy-01.korotkov.co.uk/f/8/5/6/8561c38173473cb8b42768614ffd256982.png "$$\[{\text{V}}_{\text{3}}^{\text{4}} = {\text{z}}^{\text{2}} + {\text{y}}^{\text{2}} = \left( {{\text{V}}_{\text{2}}^{\text{4}} + {\text{V}}_{\text{1}}^{\text{4}} } \right)^2 + \left( {{\text{V}}_{\text{2}}^{\text{4}} - {\text{V}}_{\text{1}}^{\text{4}} } \right)^2 = 2\left( {{\text{V}}_{\text{2}}^{\text{8}} + {\text{V}}_{\text{1}}^{\text{8}} } \right).\]$$")
![$$\[{\text{V}}_{\text{3}}^{\text{4}} = \left( {{\text{2V}}_{\text{3}}^{\text{'}} } \right)^4 = 2\left( {{\text{V}}_{\text{2}}^{\text{8}} + {\text{V}}_{\text{1}}^{\text{8}} } \right),{\text{ V}}_{\text{3}}^{{\text{'4}}} = \frac{{{\text{V}}_{\text{2}}^{\text{8}} + {\text{V}}_{\text{1}}^{\text{8}} }}{{2^3 }}.\]$$](https://dxdy-01.korotkov.co.uk/f/0/9/6/0966430593bcb9f50d6a7a7472388ed482.png "$$\[{\text{V}}_{\text{3}}^{\text{4}} = \left( {{\text{2V}}_{\text{3}}^{\text{'}} } \right)^4 = 2\left( {{\text{V}}_{\text{2}}^{\text{8}} + {\text{V}}_{\text{1}}^{\text{8}} } \right),{\text{ V}}_{\text{3}}^{{\text{'4}}} = \frac{{{\text{V}}_{\text{2}}^{\text{8}} + {\text{V}}_{\text{1}}^{\text{8}} }}{{2^3 }}.\]$$")
Чётность ![$\[\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} -\]$](https://dxdy-04.korotkov.co.uk/f/f/c/5/fc5f9506d25a02f9a9dfb826d46a18df82.png "$\[\left( {{\text{V}}_{\text{1}} {\text{,V}}_{\text{2}} {\text{ }}} \right) = {\text{d}} = {\text{1}} -\]$")
различная, и ![$\[{\text{V}}_{\text{3}}^{{\text{'4}}} \]$](https://dxdy-01.korotkov.co.uk/f/c/1/9/c19f00e080a1d333299ca5e26bee1deb82.png "$\[{\text{V}}_{\text{3}}^{{\text{'4}}} \]$")
дробное число. Это исключает разрешимость уравнения согласно требованиям при значении степени 
![$$x_1^4 = \left( {\frac{{V_2^4 + V_1^4 - 1}}
{2} - \frac{{V_2^4 - V_1^4 - 1}}
{2}} \right)\left( {\frac{{V_2^4 + V_1^4 - 1}}
{2} + \frac{{V_2^4 - V_1^4 - 1}}
{2} + 1} \right)\left[ {\left( {V_2^4 + V_1^4 } \right)^2 + \left( {V_2^4 - V_1^4 } \right)^2 } \right] = $$](https://dxdy-01.korotkov.co.uk/f/8/b/e/8bed5166973244526fbf53caa97ee48e82.png "$$x_1^4 = \left( {\frac{{V_2^4 + V_1^4 - 1}}
{2} - \frac{{V_2^4 - V_1^4 - 1}}
{2}} \right)\left( {\frac{{V_2^4 + V_1^4 - 1}}
{2} + \frac{{V_2^4 - V_1^4 - 1}}
{2} + 1} \right)\left[ {\left( {V_2^4 + V_1^4 } \right)^2 + \left( {V_2^4 - V_1^4 } \right)^2 } \right] = $$")
, 
, 
, 
, 
, 
.
(и потом постоянно на неё ссылаетесь), ![$\[f\left( {y,z} \right)\]$](https://dxdy-02.korotkov.co.uk/f/5/7/4/574912faa2dd3afb2e9ad42dcd99199182.png "$\[f\left( {y,z} \right)\]$")
(и потом постоянно на неё ссылаетесь), не потрудившись объяснить, что это за функция.![$\[f\left( {y,z,...} \right)\]$](https://dxdy-04.korotkov.co.uk/f/7/5/3/753952c87672538df305032c01f88a0082.png "$\[f\left( {y,z,...} \right)\]$")
, собственно её степени (где ![$\[x^2 = \left( {z - y - w} \right)\left( {z + y + w} \right) = U_1^2 U_2^2 - \]$](https://dxdy-03.korotkov.co.uk/f/6/2/a/62a0adb3e8bd9b11ed0b476c6b88b97282.png "$\[x^2 = \left( {z - y - w} \right)\left( {z + y + w} \right) = U_1^2 U_2^2 - \]$")
неопределённое уравнение,![$$\[\left\{ \begin{gathered} z - y - w = U_{\text{1}}^2 \hfill \\ z + y + w = U_{\text{2}}^2 {\text{, }} \hfill \\ \end{gathered} \right.z = \frac{{U_{\text{2}}^2 + U_{\text{1}}^2 }}{2},y = \frac{{U_{\text{2}}^2 - U_{\text{1}}^2 - 2w}}{2},(U_1 ,U_2 ) = d = 1 - \]$$](https://dxdy-03.korotkov.co.uk/f/2/7/e/27e4a53edf9b847d2563c8254cc1a78982.png "$$\[\left\{ \begin{gathered} z - y - w = U_{\text{1}}^2 \hfill \\ z + y + w = U_{\text{2}}^2 {\text{, }} \hfill \\ \end{gathered} \right.z = \frac{{U_{\text{2}}^2 + U_{\text{1}}^2 }}{2},y = \frac{{U_{\text{2}}^2 - U_{\text{1}}^2 - 2w}}{2},(U_1 ,U_2 ) = d = 1 - \]$$")
![$\[w - \]$](https://dxdy-01.korotkov.co.uk/f/0/6/c/06c567f45bfa5dd8f6b6cfc8ddbefb0882.png "$\[w - \]$")
подбирается произвольно,![$\[x^3 = \left( {z - y - w} \right)\left( {z + y + w} \right)\left( {z - y + w} \right) = U_1 U_2 U_3 - \]$](https://dxdy-04.korotkov.co.uk/f/7/0/4/704682413d2304396739dfcc1d1de72782.png "$\[x^3 = \left( {z - y - w} \right)\left( {z + y + w} \right)\left( {z - y + w} \right) = U_1 U_2 U_3 - \]$")
определённое уравнение,![$$\[\left\{ \begin{gathered} z - y - w = U_1 \hfill \\
z + y + w = U_{\text{2}} \hfill \\ z - y + w = U_{\text{3}} {\text{, }} \hfill \\ \end{gathered} \right.z = \frac{{U_2 + U_1 }}
{2},y = \frac{{U_2 - U_3 }}{2},w = \frac{{U_3 - U_1 }}
{2},(U_1 ,U_2 ,U_3 ) = d = 1-\]$$](https://dxdy-03.korotkov.co.uk/f/e/e/d/eed86bf867ab25651d8573620288ba2382.png "$$\[\left\{ \begin{gathered} z - y - w = U_1 \hfill \\
z + y + w = U_{\text{2}} \hfill \\ z - y + w = U_{\text{3}} {\text{, }} \hfill \\ \end{gathered} \right.z = \frac{{U_2 + U_1 }}
{2},y = \frac{{U_2 - U_3 }}{2},w = \frac{{U_3 - U_1 }}
{2},(U_1 ,U_2 ,U_3 ) = d = 1-\]$$")
![$\[x^3 = U_1 U_2 U_3 ,\]$](https://dxdy-02.korotkov.co.uk/f/1/d/a/1dad0166545d1dea6ab0849abcef389f82.png "$\[x^3 = U_1 U_2 U_3 ,\]$")
![$ \[x^2 = \left( {z - y} \right)\left( {z + y} \right)\left( {z - y - 1} \right) = U_1 U_2 U_3 - \]$](https://dxdy-02.korotkov.co.uk/f/5/3/3/5335e121d06bd4ffd0a6ac128a4f937f82.png "$ \[x^2 = \left( {z - y} \right)\left( {z + y} \right)\left( {z - y - 1} \right) = U_1 U_2 U_3 - \]$")
переопределённое уравнение,![$$\[\left\{ \begin{gathered} z - y = U_1 \hfill \\ z + y = U_2 \hfill \\
z - y - 1 = U_3 {\text{, }} \hfill \\ \end{gathered} \right.z = \frac{{U_2 + U_1 }}{2},y = \frac{{U_2 - U_1 }}{2},x^2 = U_1 U_2 U_3 .\]$$](https://dxdy-01.korotkov.co.uk/f/0/e/3/0e370741f2a97715f603a19007da02a982.png "$$\[\left\{ \begin{gathered} z - y = U_1 \hfill \\ z + y = U_2 \hfill \\
z - y - 1 = U_3 {\text{, }} \hfill \\ \end{gathered} \right.z = \frac{{U_2 + U_1 }}{2},y = \frac{{U_2 - U_1 }}{2},x^2 = U_1 U_2 U_3 .\]$$")
![$$\[U_3 = z - y - 1 = \frac{{U_2 + U_1 }}{2} - \frac{{U_2 - U_1 }}{2} - 1 = U_1 - 1,U_3 = U_1 - 1.\]$$](https://dxdy-02.korotkov.co.uk/f/5/6/5/565abf3a49206798291c88ef27c218ab82.png "$$\[U_3 = z - y - 1 = \frac{{U_2 + U_1 }}{2} - \frac{{U_2 - U_1 }}{2} - 1 = U_1 - 1,U_3 = U_1 - 1.\]$$")
Исхоное уравнение имеет решение, ибо уравнение ![$\[U_3 = U_1 - 1\]$](https://dxdy-01.korotkov.co.uk/f/8/9/8/89807a5a8b021fcb87e098b29aba3ada82.png "$\[U_3 = U_1 - 1\]$")
разрешимо в целых числах, при условии ![$ \[x^2 = U_1 U_2 U_3 .\]$](https://dxdy-03.korotkov.co.uk/f/e/f/1/ef1f6606d978ecf410beacd387f2002a82.png "$ \[x^2 = U_1 U_2 U_3 .\]$")
![$\[U_1 ,\]$](https://dxdy-01.korotkov.co.uk/f/0/3/d/03dd63bf8a9fead2a273f4d7b32b62c182.png "$\[U_1 ,\]$")
например ![$\[U_1 = 4,\]$](https://dxdy-02.korotkov.co.uk/f/1/5/f/15ff5fe0cf6388c3dc837b282f6cdac682.png "$\[U_1 = 4,\]$")
вычисляя значение ![$\[U_3 \]$](https://dxdy-02.korotkov.co.uk/f/1/4/b/14bdecec276c00a4bc8a821ddb6bcc3c82.png "$\[U_3 \]$")
и подбирая значение ![$\[U_2 ,\]$](https://dxdy-01.korotkov.co.uk/f/8/3/0/8303876155c2e9cc354f6b573bbe58c282.png "$\[U_2 ,\]$")
имеем:![$$\[x^2 = U_1 U_2 U_3 = 4 \cdot U_2 \cdot 3 = 4 \cdot 12 \cdot 3 = 12^2 ,x = 12,y = 4,z = 3.\]$$](https://dxdy-03.korotkov.co.uk/f/2/a/3/2a3cabbfe94768fb7e2b1ba2b57bcb6082.png "$$\[x^2 = U_1 U_2 U_3 = 4 \cdot U_2 \cdot 3 = 4 \cdot 12 \cdot 3 = 12^2 ,x = 12,y = 4,z = 3.\]$$")
, собственно её степени (где 
- значение степени