Determining number of integer solutions for expression of perfect square How many positive integer values of n are there such that $2^n + 7^n$ is a perfect square?


I am not sure how to approach this question given that there are two different bases 2 and 7
The only instance is $n=1$.

Let $a_n=2^n+7^n$.

Of course $a_1$ is a square.  Assume that $n&gt;1$.

We see that $a_n\equiv (-1)^n \pmod 4$ so $n$ must be even (this is where we use $n&gt;1$). 

Similarly we see that $a_n\equiv 2^{n+1}\pmod 5$ which implies that $n$ is odd, and we are done
This the first attempted solution not only on this post but also my overall first.

N can take either an even value or an odd value.
Im attempting only one part-easier one.

Case I(N is even) 

The last digit of powers of 2 and 7 repeat in groups of four i.e (2,4,8,6) and (7,9,3,1).For N=4k+2 and N=4k both the last digit of the overall expression turns out to be (4+9=)3 and (6+1=)7,both of which can never be last digits of a perfect square.

(or)

Simply take modulo 3 when n is even,which tells us that the whole expression is 2(mod 3),never a perfect square.

Case II-N is odd
When n is odd and greater than 1(for which prob. statement holds true), taking modulo 4 gives a remainder of 3,neither 1 nor 0,hence it wont be a perfect square.