another intresting equation $p^x$ + $q^y$ -$r^z$ = 0

members, the equation $p^x$ + $q^y$ - $r^z$ = 0 (where r is odd integer) has only positive integr solutions iff the following conditions made.

a) p = -1 (mod q^(2k)), here k is any positive integer.

b) 4|q -3

c) $p^2$ + q^(2k-1) = r

Then our equation $p^x$ + $q^y$ -$r^z$ = 0 has positive solutions in the form of (2, 2k-1, 1).

I am almost did. But, I need little more discussion to reach this solution and I want to know that if r is not odd, how to guess the solutions.

Thanks in advance to all members of stack exchange.

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 I'm not exactly sure what is being asked, but observe that $2^5+7^2-3^4=0$ which doesn't match your conditions and is a solution for $z=2$ and $z=4$. – Zander Apr 8 '12 at 12:33 Zander Sir! as per your equation p = 2, q = 7 and r = 3 – gandhi Apr 8 '12 at 14:55 if we see condition (c) of my post, $p^2$ + q^(2k-1) = r. According to Zander, it is failed for any k. Please check. – gandhi Apr 8 '12 at 14:56 I am looking solutions as per my all a, b and c conditions. Then, we get solutions in the form of (2, 2k-1, 1). – gandhi Apr 8 '12 at 14:57