Day 8: The Power of Eight

First, as the RHS is an integer, $a$ and $b$ must be non-negative. Now we work in $\bmod 8$ .

For $a = 0, 1, 2$ , $2^a \equiv 1, 2, 4 \bmod 8$ respectively, then for $a > 2$ , $2^a \equiv 0 \bmod 8$ . Also, $7^b \equiv 1, 7 \bmod 8$ .

Case $a > 2$ : The LHS can only be $1,7 \bmod 8$ but the RHS can only be $4,5 \bmod 8$ hence there are no solutions for $a > 2$ .

Case $a = 0$ : Now we are left with $n^2 = 7^b + 5$ . If $b=0$ then we obviously have no solution, and if $b > 0$ , then $n^2 \equiv 5 \bmod 7$ which is not possible. Hence there are no solutions for $a=0$ .

Case $a = 1$ : Now we are left with $n^2 = 7^b + 6$ and the same argument as above shows that there are no solutions for $a=1$ .

Case $a = 2$ : Now we are left with $n^2 = 7^b + 8$ . We see that $7^b \equiv 1 \bmod 8$ and not $7 \bmod 8$ as the LHS is a square. We conclude that $b$ is even, so we write it as $2k$ for some integer $k$ . Now we can rewrite the equation as follows:

$8 = (n + 7^k)(n - 7^k)$

By considering the pairs of factors of $8$ , we conclude that the only solutions are $(2, 0, 3)$ , $(2, 0, -3)$ for $a = 2$ .

Therefore the only solutions are $(2, 0, 3)$ , $(2, 0, -3)$ and hence the answer is $2 + 2 = \boxed{4}$ .

$a,b \ge 0$ , $|n| \ge 3$ , let's limit ourselves to positive $n$ . Suppose that $a \ge 3$ . Modulo 8, we get $\left\lbrace 1,7\right\rbrace \equiv (-1)^b \equiv n^2-4 \equiv \left\lbrace 4,5\right\rbrace$ , which clearly can't be satisfied. Therefore, $a \le 2$ . This is the obligatory (and hinted at) step.

Modulo 3, we get $\left\lbrace 0,2\right\rbrace \equiv (-1)^a+1^b \equiv n^2-1 \equiv \left\lbrace 0,7\right\rbrace$ . This is satisfied only for both congruences equal to 0, which implies $n$ divisible by 3 and $b$ even.

Now, let's rewrite the original equation as $n^2-7^b=(n-7^{b/2})(n+7^{b/2})=4+2^a$ . The left side needs to be positive, so $n \ge 7^{b/2}$ and therefore $\left\lbrace5,6,8\right\rbrace=4+2^a=n^2-7^b \ge 2\cdot7^{b/2}+1$ . For $b/2 \ge 1$ , the right side can't be $\le 8$ , so there's no solution. For $b/2=0$ , $n \ge 4$ makes $n^2-7^b=n^2-1$ larger than 8, so that won't work either, Since $n$ has to be divisible by 3, the only solution for positive $n$ is $3$ ; it's paired with an analogous solution for $n=-3$ . Both of them have $a=2, b=0$ , so the answer is $2+0+3+2+0-3=4$ .