Determine all possible positive integer(s) P such that:

2¹⁹⁹⁴ + 2¹⁹⁹⁸ + 2¹⁹⁹⁹ + 2²⁰⁰⁰ + 2²⁰⁰² + 2^P

is a perfect square.

2^1994(1 + 2^4 + 2^5 + 2^6 ) + 2^p is a perfect square

That is, 2^1994(113) + 2^p is a perfect square

So, 2^1994(113 + 2^p-1994) is a perfect square

Since 2^1994 is a perfect square, we need the quantity in parentheses also to be a perfect square.

This occurs when p – 1994 = 3, which makes the quantity in parentheses = 121

Therefore, p = 1997
which appears to be the only positive integral solution

Edited on July 8, 2009, 5:32 pm

Posted by JayDeeKay on 2009-07-08 17:29:57