
Отсюда

Примем

за аргумент:

Интересные соотношения получаются.

и

Видим, что

, т.е. двойка есть квадратичный вычет по модулю

.
Тогда любой простой делитель

, назовём его

, должен иметь символ Лежандра равный единице

, и соответственно символ Кронекера

.
Следующий небольшой тест это подтверждает.
Код:
testd()=
{
for(l=2, 100,
if(!issquare(l+1),
for(y=2, sqrtint(l),
d= gcd(l+1, y^2-2);
if(d>1, if(issquarefree(d),
print("l = "l" d = "d" "factorint(d))
))
)
)
)
};
(вывод теста)
? \r testd.gp
? testd()
l = 5 d = 2 Mat([2, 1])
l = 7 d = 2 Mat([2, 1])
l = 9 d = 2 Mat([2, 1])
l = 11 d = 2 Mat([2, 1])
l = 13 d = 2 Mat([2, 1])
l = 13 d = 7 Mat([7, 1])
l = 17 d = 2 Mat([2, 1])
l = 17 d = 2 Mat([2, 1])
l = 19 d = 2 Mat([2, 1])
l = 19 d = 2 Mat([2, 1])
l = 20 d = 7 Mat([7, 1])
l = 20 d = 7 Mat([7, 1])
l = 21 d = 2 Mat([2, 1])
l = 21 d = 2 Mat([2, 1])
l = 23 d = 2 Mat([2, 1])
l = 23 d = 2 Mat([2, 1])
l = 25 d = 2 Mat([2, 1])
l = 25 d = 2 Mat([2, 1])
l = 27 d = 2 Mat([2, 1])
l = 27 d = 7 Mat([7, 1])
l = 27 d = 14 [2, 1; 7, 1]
l = 29 d = 2 Mat([2, 1])
l = 29 d = 2 Mat([2, 1])
l = 31 d = 2 Mat([2, 1])
l = 31 d = 2 Mat([2, 1])
l = 33 d = 2 Mat([2, 1])
l = 33 d = 2 Mat([2, 1])
l = 34 d = 7 Mat([7, 1])
l = 34 d = 7 Mat([7, 1])
l = 37 d = 2 Mat([2, 1])
l = 37 d = 2 Mat([2, 1])
l = 37 d = 2 Mat([2, 1])
l = 39 d = 2 Mat([2, 1])
l = 39 d = 2 Mat([2, 1])
l = 39 d = 2 Mat([2, 1])
l = 41 d = 2 Mat([2, 1])
l = 41 d = 7 Mat([7, 1])
l = 41 d = 14 [2, 1; 7, 1]
l = 41 d = 2 Mat([2, 1])
l = 43 d = 2 Mat([2, 1])
l = 43 d = 2 Mat([2, 1])
l = 43 d = 2 Mat([2, 1])
l = 45 d = 2 Mat([2, 1])
l = 45 d = 2 Mat([2, 1])
l = 45 d = 23 Mat([23, 1])
l = 45 d = 2 Mat([2, 1])
l = 47 d = 2 Mat([2, 1])
l = 47 d = 2 Mat([2, 1])
l = 47 d = 2 Mat([2, 1])
l = 49 d = 2 Mat([2, 1])
l = 49 d = 2 Mat([2, 1])
l = 49 d = 2 Mat([2, 1])
l = 50 d = 17 Mat([17, 1])
l = 51 d = 2 Mat([2, 1])
l = 51 d = 2 Mat([2, 1])
l = 51 d = 2 Mat([2, 1])
l = 53 d = 2 Mat([2, 1])
l = 53 d = 2 Mat([2, 1])
l = 53 d = 2 Mat([2, 1])
l = 55 d = 2 Mat([2, 1])
l = 55 d = 7 Mat([7, 1])
l = 55 d = 14 [2, 1; 7, 1]
l = 55 d = 2 Mat([2, 1])
l = 57 d = 2 Mat([2, 1])
l = 57 d = 2 Mat([2, 1])
l = 57 d = 2 Mat([2, 1])
l = 59 d = 2 Mat([2, 1])
l = 59 d = 2 Mat([2, 1])
l = 59 d = 2 Mat([2, 1])
l = 61 d = 2 Mat([2, 1])
l = 61 d = 2 Mat([2, 1])
l = 61 d = 2 Mat([2, 1])
l = 62 d = 7 Mat([7, 1])
l = 62 d = 7 Mat([7, 1])
l = 65 d = 2 Mat([2, 1])
l = 65 d = 2 Mat([2, 1])
l = 65 d = 2 Mat([2, 1])
l = 65 d = 2 Mat([2, 1])
l = 67 d = 2 Mat([2, 1])
l = 67 d = 2 Mat([2, 1])
l = 67 d = 34 [2, 1; 17, 1]
l = 67 d = 2 Mat([2, 1])
l = 68 d = 23 Mat([23, 1])
l = 69 d = 2 Mat([2, 1])
l = 69 d = 7 Mat([7, 1])
l = 69 d = 14 [2, 1; 7, 1]
l = 69 d = 2 Mat([2, 1])
l = 69 d = 2 Mat([2, 1])
l = 71 d = 2 Mat([2, 1])
l = 71 d = 2 Mat([2, 1])
l = 71 d = 2 Mat([2, 1])
l = 71 d = 2 Mat([2, 1])
l = 73 d = 2 Mat([2, 1])
l = 73 d = 2 Mat([2, 1])
l = 73 d = 2 Mat([2, 1])
l = 73 d = 2 Mat([2, 1])
l = 75 d = 2 Mat([2, 1])
l = 75 d = 2 Mat([2, 1])
l = 75 d = 2 Mat([2, 1])
l = 75 d = 2 Mat([2, 1])
l = 76 d = 7 Mat([7, 1])
l = 76 d = 7 Mat([7, 1])
l = 77 d = 2 Mat([2, 1])
l = 77 d = 2 Mat([2, 1])
l = 77 d = 2 Mat([2, 1])
l = 77 d = 2 Mat([2, 1])
l = 79 d = 2 Mat([2, 1])
l = 79 d = 2 Mat([2, 1])
l = 79 d = 2 Mat([2, 1])
l = 79 d = 2 Mat([2, 1])
l = 81 d = 2 Mat([2, 1])
l = 81 d = 2 Mat([2, 1])
l = 81 d = 2 Mat([2, 1])
l = 81 d = 2 Mat([2, 1])
l = 83 d = 2 Mat([2, 1])
l = 83 d = 7 Mat([7, 1])
l = 83 d = 14 [2, 1; 7, 1]
l = 83 d = 2 Mat([2, 1])
l = 83 d = 2 Mat([2, 1])
l = 84 d = 17 Mat([17, 1])
l = 85 d = 2 Mat([2, 1])
l = 85 d = 2 Mat([2, 1])
l = 85 d = 2 Mat([2, 1])
l = 85 d = 2 Mat([2, 1])
l = 87 d = 2 Mat([2, 1])
l = 87 d = 2 Mat([2, 1])
l = 87 d = 2 Mat([2, 1])
l = 87 d = 2 Mat([2, 1])
l = 89 d = 2 Mat([2, 1])
l = 89 d = 2 Mat([2, 1])
l = 89 d = 2 Mat([2, 1])
l = 89 d = 2 Mat([2, 1])
l = 90 d = 7 Mat([7, 1])
l = 90 d = 7 Mat([7, 1])
l = 91 d = 2 Mat([2, 1])
l = 91 d = 2 Mat([2, 1])
l = 91 d = 23 Mat([23, 1])
l = 91 d = 2 Mat([2, 1])
l = 91 d = 2 Mat([2, 1])
l = 92 d = 31 Mat([31, 1])
l = 93 d = 2 Mat([2, 1])
l = 93 d = 2 Mat([2, 1])
l = 93 d = 2 Mat([2, 1])
l = 93 d = 47 Mat([47, 1])
l = 93 d = 2 Mat([2, 1])
l = 95 d = 2 Mat([2, 1])
l = 95 d = 2 Mat([2, 1])
l = 95 d = 2 Mat([2, 1])
l = 95 d = 2 Mat([2, 1])
l = 97 d = 2 Mat([2, 1])
l = 97 d = 7 Mat([7, 1])
l = 97 d = 14 [2, 1; 7, 1]
l = 97 d = 2 Mat([2, 1])
l = 97 d = 2 Mat([2, 1])
Видно, что, кроме двойки, в кандидатах

простыми делителями являются только
7,17,23,31,47, у которых символ Лежандра равен единице.
Возможно ли это как-то использовать в поиске контрпримера? И что можно сказать о чётности

?