(In reply to Possible solution
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 x2 - 5y2 = 1 becomes 4X2 - 5y2 = 1.
Then working modulo 4, this gives y2 = 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 2010-12-03 21:26:03