(In reply to
Possible solution by broll)
I agree with Jer; your 'elementary' method is interesting and worth perfecting, but the 'oddness' of x still needs to be proven. Would this do the trick?...
Assuming x is even, write x = 2X, so that x^{2}  5y^{2} = 1 becomes 4X^{2}  5y^{2} = 1.
Then working modulo 4, this gives y^{2} = 3 (mod 4).
Now, squares can only take the values 0 or 1 (mod 4), so we have a contradiction, proving that if a solution exists for this equation, then x is odd.
Edited on December 3, 2010, 9:28 pm

Posted by Harry
on 20101203 21:26:03 