Equation in natural numbers

[ins-]

Solve in natural numbers the equation: $1=\frac{2}{x^2}+\frac{3}{y^2}+\frac{4}{z^2}$.

[denisart]

ins- в сообщении #1486332 писал(а):

Solve in natural numbers the equation: $1=\frac{2}{x^2}+\frac{3}{y^2}+\frac{4}{z^2}$.

Из неравенства Коши-Шварца
$$1 \le 29 \left(\dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}\right)$$.

Из последнего неравенства хотя бы одна из переменных должна быть меньше $4$-х. Далее простой перебор.

$x=3, y=3, z=3$.

[ins-]

$x=y=z=3$ is not the only solution.

[nnosipov]

Yes, we have also $x=3$, $y=2$, $z=12$. However, the problem is not intriguing. What is interesting: the similar equation $1=2/x^2-3/y^2+4/z^2$ seems to be unsolvable in positive integers.

[denisart]

nnosipov в сообщении #1486353 писал(а):

Yes, we have also $x=3$, $y=2$, $z=12$. However, the problem is not intriguing. What is interesting: the similar equation $1=2/x^2-3/y^2+4/z^2$ seems to be insolvable in positive integers.

Да, это решение во время перебора упустил. Спасибо.

[ins-]

nnosipov в сообщении #1486353 писал(а):

Yes, we have also $x=3$, $y=2$, $z=12$. However, the problem is not intriguing. What is interesting: the similar equation $1=2/x^2-3/y^2+4/z^2$ seems to be insolvable in positive integers.

Is it possible to use here: $2/x^2+4/z^2=1+3/y^2>1$ then finding some bounds for $x$ or $z$ and checking cases? The initial equation was proposed in some Bulgarian competitions - 1981-st and then 1996-th - municipality round.

[nnosipov]

ins- в сообщении #1486359 писал(а):

Is it possible to use here: ...

Yes, you are right. This is the same.