$ 7^{2n} -48n - 1=4^n-3n-1 \pmod{9}$ Let $n=3k+m$ where $m=0,1,2$ then $4^n-3n-1=4^{3k+m}-9k-3m-1=4^m-3m-1 \pmod{9}$ so we need to check only $m=0,1,2$, which is easy.

For $n=1$ it's obvious. Just note that $$A_{n+1}=49\cdot A_n+48^2\cdot n\equiv_94\cdot A_n$$The rest is trivial induction.

Yep, induction is the way to go.

$7^{2n} - 48n - 1 \equiv 49^n - 3n - 1 \equiv 4^n - 3n - 1$(mod 9). Now, we can plug all residues $mod$ $9$ and we will se this is always $0 \equiv$ (mod 9)

let n=k+1 +48k*49-48k*49

Last year I gave the following generalization in the JBMO Group at the Bulgarian Festival of Young Mathematicians at Sozopol in September. Find all pairs $(a,n)$ of positive integers, for which $a^{2n} + (1-a^2)n - 1 $ is divisible by $9$. Answer. $(3k, 9\ell+1)$, $(3k-1,m)$, $(3k-2,m)$ for arbitrary positive integers $k$, $\ell$, $m$ Solution. If $a$ is a multiple of $3$, we immediately get $n\equiv 1 \pmod 9$. Suppose $a$ is not a multiple of $3$ - then direct verification shows (e.g. by computing with $a\equiv \pm 1, \pm 2, \pm 4$), that $a^6 \equiv 1 \pmod 9$, hence $a^{2n} \equiv 1, a^2, a^4 \pmod 9$ for $n\equiv 0,1,2 \pmod 3$, respectively. If $n$ is a multiple of $3$, then $a^{2n} - 1$ is divisible by $9$, while $(1-a^2)n = (1-a)(1+a)n$ also, since $a$ is not divisible by $3$, i.e. exactly one of $1-a$ and $1+a$ is divisible by $3$. If $n\equiv 1 \pmod 3$, the expression is congruent to $a^2 + (1-a^2)n - 1 = (1-a)(1+a)(n-1)$ and is divisible by $9$, as in the previous case. If $n\equiv 2 \pmod 3$, the expression is congruent to $a^4 + (1-a^2)n - 1 = (a-1)(a+1)(a^2+1-n)$ and is divisible by $9$, since exactly one of $a-1$ and $a+1$ is divisible by $3$, while $n$ and $a^2+1$ give remainder $2$ when divided by $3$.

Viktor_Maks wrote: $7^{2n} - 48n - 1 \equiv 49^n - 3n - 1 \equiv 4^n - 3n - 1$(mod 9). Now, we can plug all residues $mod$ $9$ and we will se this is always $0 \equiv$ (mod 9) I would like to mention something like this is not generally true. Take for example the problem prove that for every positive integer $$2^n + 2x^4 + 3x^3 + x^2 + 4x + 3$$is divisible by $5$. Just because you plugged $1 , 2 , 3 , 4 , 5$ and got that in all those cases it is indeed divisible by $5$ doesnt mean it is divisible by $5$. As we can see that expression is typically not divisible by $5$. So what happened? Well the whole logic behind this plugging $x$ values and if its true its always true comes from the fact that polynomials $\pmod n$ are periodic with period $n$ (it could theoretically be even some divisor of $n$ but lets ignore that for now) . But exponentials $\pmod n$ are periodic with period $ord_n(a)$ so when it comes to period of the sum it is $lcm(n , ord_n(a))$ now in this exact problem it just happens that $lcm$ is indeed $9$ but still. In my eyes something like that is a fake solve as nobody checked if checking only residues $\pmod 9$ is enough. (Correct me if im wrong) In short words $a \equiv b \pmod n$ Doesnt imply $c^a \equiv c^b \pmod n$