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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

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  • $\begingroup$ Reduce this sequence modulo a suitable quantity. $\endgroup$ – Allawonder Sep 11 at 12:50
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The only instance is $n=1$.

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

Of course $a_1$ is a square. Assume that $n>1$.

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

Similarly we see that $a_n\equiv 2^{n+1}\pmod 5$ which implies that $n$ is odd, and we are done

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Set $2^n+7^n=x^2$. Since all perfect squares are congruent to $0$ or $1$ modulo $3$ . On the other hand $7^n\equiv 1\pmod 3$ and $2^n\equiv 1\pmod 3$ or $2^n\equiv 2\pmod 3$ .This means that $n$ must be odd(why?). Proof by Contradiction: We wish to show that the only value of $n$ is $1$. Assume that $n>2$. Then consider the equation $2^n=x^2-7^n$. let $x=2a+1$. We have $2^n=4a^2+4a+1-7^n$. Dividing by $4$, $2^ {n-2}=a^2+a+(1-7^n)/4$. the entire LHS is an integer, and so are $a^2$ and $a$. Thus, $\dfrac {1}{4} (1 - 7^n)$ must be an integer. Let $\dfrac {1}{4} (1 - 7^n) = k$.

Then we have $1- 7^n = 4k$. This means that $1- 7^n \equiv 0 \pmod {4}$. Thus, $n$ is even. However, it has already been shown that $n$ must be odd. This is a contradiction. Therefore, $n$ is not greater than or equal to $ 2$, and must hence be less than $2$. The only positive integer less than $2$ is $1$

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If $n=2m$, $2^{2m}+7^{2m}$ is too close to the square of $7^m$ to be a square itself. This gives that $n$ is odd.
If we consider $n=2m+1$ for $m\geq 1$ (the case $n=1$ leads to a trivial solution) we have that $$ 2^n+7^n = 2\cdot 4^m + (8-1)^{2m+1} \equiv (-1)^{2m+1} \equiv -1\pmod{4}, $$ so $2^n+7^n$ cannot be a square (a square $\!\!\pmod{4}$ is either $0$ or $1$).
It follows that the trivial solution is the only solution.

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This the first attempted solution not only on this post but also my overall first:I can only give you a start with this answer.

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.

Ill try to complete it soon.(Sorry for not using LaTex as my knowledge of it is non existent.)

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