Positive Integers to make a square

Question

How many integers $n$ make the expression $7^n + 7^3 + 2\cdot7^2$ a perfect square?

Factoring $7^2$ we have that $7^n + 7^3 + 2\cdot7^2 = 7^2\cdot(7^{n-2} + 9)$.
How do we prove that the 2nd factor only has $1$ solution when $n = 3$?

Answer 1

An "equivalent" way of doing things, is to note that $7^3 + 2 \times 7^2 = 343+98=441 = 21^2$.

Now, if $7^n + 21^2 = y^2$ for some $y$, then $7^n = (y-21)(y+21)$.

So the prime factorization of $(y-21)$ and $(y+21)$ can both contain only the prime $7$, by unique factorization theorem. Therefore, $y - 21$ and $y+21$ are both powers of $7$. The difference between these powers of $7$ is $y+21 - (y-21) = 42$.

Only two such powers have a difference of $42$, which are $49$ and $7$.This is not very difficult to see from the fact that $7^n - 7^m = 7^{n-m}(7^{m} - 1) = 42$ forces $n-m = 1$ and $m = 1$. Therefore, $(y-21)(y+21) = 7 \times 49 = 343$, so $n = 3$ is forced from here.

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Thinking of it, you could also have done $7^{n-2} + 9 = x^2$ since you factorized out $7^2$, but the same logic will apply, since $9 = 3^2$.

Answer 2

Given $7^n+7^3+2\cdot7^2$

Note that $7^3+2\cdot7^2=441=21^2$

So, our equation becomes, $$7^n+441=p^2\mbox{ for some integer p} $$$$7^n=p^2-441$$$$7^n=(p+21)(p-21)$$

Since $7$ is a prime number, $(p+21)$ and $(p-21)$ should be degrees of $7$

So, the only possible value of $n$ is $3$