While I was glancing at my old books, I saw a question.
Let $A=n^2+48$ and $n<500$. Then find the whole satisfied values for $n$ when $A$ is divided by $49$, the remainder is zero.
I thought that if $n^2 \equiv -48 \pmod{49}$ then $n^2 \equiv 1\pmod{49}$. However I'm stuck here. What can I do to find all $n$ values?
you can do this one degree of 7 at a time. For $n^2 \equiv 1 \pmod 7,$ we have $1,6 \pmod 7.$ That is already fairly restrictive. next, we try one up, unknown $s$ with $$ ( 1 + 7 s)^2 \equiv 1 \pmod {7^2} $$ Well, $$ ( 1 + 7 s)^2 = 1 + 14 s + 49 s^2 \equiv 1 + 14 s \equiv 1 \pmod {7^2} $$ so that $s $ is zero, actually from $s \equiv 0 \pmod 7.$
Next, unknown $t$ with $$ ( 6 + 7 t)^2 \equiv 1 \pmod {7^2} $$ Well, $$ ( 6 + 7 t)^2 = 36 + 84 t + 49 t^2 \equiv 36 + 84 t\equiv 36 + 35 t \equiv 1 \pmod {7^2} $$ so that $t \equiv -1 \equiv 6 \pmod 7.$ That makes this value $6 + 7t \equiv 6 + 42 \equiv 48 \equiv -1 \pmod {7^2}$
That is, there are still just two square roots, $\pm 1 \pmod {49}$
If $49$ divides $(n-1)(n+1)$, then so does $7$. Since $7$ is prime, this implies that $7$ divides $n-1$ or $n+1$. If $7$ divides $n-1$, then $7$ cannot divide $n+1$, and so $7^2$ must divide $n-1$. Likewise for $n+1$.
Bottom line: $n^2 \equiv 1 \bmod 49$ iff $n \equiv \pm1 \bmod 49$.
This argument works for all odd primes $p$: $n^2 \equiv 1 \bmod p^2$ iff $n \equiv \pm1 \bmod p^2$.
One brute force solution would be to try finding the quadratic residues when you square each number from $1$ to $24$. The congruence class of $n^2$ mod $49$ is completely determined by the congruence class of $n$. Additionally, you don't have to check the classes from $25$ to $48$ because, for example, $$ 37\cong -12 (\textrm{mod} 9) $$ and so we know that $$ 37^2\cong (-12)^2\cong12^2 (\textrm{mod} 9). $$ So immediately we get that any number of the form $-1,1(\textrm{mod} 49)$ squares to $1$.
@WillJagy's solution is a bit more elegant, and the suggestion from @cgss looks like the best way.
