squares

[christi]

Find the squares of the form 2^n + 7^n + 27, n>0 integer.
Cristinel Mortici

[grobber]

For n=1 we get 36, which is a perfect square. If n>=2 then 2^n=M4. If n is odd then 7^n is M4+3 and 27 is M4+3, so 2^n+7^n+27 is M4+2, so it can't be a perfect square. This means that if n>=2 then n is even.

We have 27=M3, 2^n=M3+1 (because n is even) and 7^n=M3+1, so 2^n+7^n+27 is M3+2, so it can't be a perfect square.

Therefore, the only solution is n=1.

[christi]

If n is even, then the given number lies between two consecutive squares: 7^n and the next.
But your solution is easier and avoids the above remark.

PS. In fact this problems wants to be related with the following:

Find the squares of the form: n^4 + 7^n +47 (Gazeta Matematica)

[grobber]

That last problem you mentioned is quite famous by now. It's even been discussed on thids forum a couple of times

I think this topic belongs to the other "Proposed Problems" section, however. It doesn't look like college math