An evenly even number is a number in the form of 2^m , where m is a positive integer.

Prove that it is impossible the sum of two evenly even numbers to be a perfect square.

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- Sep 8th 2009, 08:53 PMnh149even numbers
An evenly even number is a number in the form of 2^m , where m is a positive integer.

Prove that it is impossible the sum of two evenly even numbers to be a perfect square. - Sep 8th 2009, 09:23 PMMonkey D. Johnny
but 2^3+2^3=16 is a perfect square

- Sep 9th 2009, 04:22 AMBruno J.
As is $\displaystyle 2^5+2^2=6^2$. Not a very good conjecture!

By the way the real name for an "evenly even number" is*power of two*. - Sep 9th 2009, 06:26 AMMonkey D. Johnny
In fact, I believe 2^m+2^n is a perfect square if and only if either:

i) both m and n are odd and m=n, or

ii)either m or n is even (say m is even) and n=m+3.

So, the example I gave was part (i) and the example Bruno J. gave was part (ii).

Proof.

The "if" part is pretty straight forward.

So assume 2^m+2^n is a perfect square. And assume part i) doesn't hold. So we will show part ii) holds. Suppose, m $\displaystyle \leq n$.

Then 2^m+2^n=2^m(1+2^(n-m)). This implies m is even and 1+2^(n-m) is a perfect square. Write 1+2^(n-m)=k^2. Then 2^(n-m)=(k-1)(k+1). This implies k-1 is a power of 2. Write k-1=2^p. Then 2^(n-m)=2^(2p)+2^(p+1). The right hand side of the last equality can be a power of 2 only if p=1. This implies 2^(n-m)=8 or n=m+3.