Hepta-tuples

Applying the AM-GM inequality, we have

$1+a^8+b^8+c^8+d^8+e^8+f^8+g^8\ge8\sqrt[8]{a^8b^8c^8d^8e^8f^8g^8}=8|abcdefg|\ge8abcdefg$

The equality holds if and only if $|a|=|b|=|c|=|d|=|e|=|f|=|g|=1$ and $abcdefg=1$

With each sextuple of $(a,b,c,d,e,f,g)$ inwhich $|a|=|b|=|c|=|d|=|e|=|f|=1$ , there is only one value of $g$ such that $abcdefg=1$ .

So $k=2^6=64$ and $k-1=63$ .

Relevant wiki: Classical Inequalities

By AM_GM $1+a^8+b^8+c^8+d^8+e^8+f^8+g^8\ge8(a^8b^8c^8d^8e^8f^8g^8)^{\frac{1}{8}}=8|abcdefg|$

Equality holds iff $|a|=|b|=|c|=|d|=|e|=|f|=|g|=1$

Now since LHS is positive we must have RHS positive. $abcdefg$ would be positive if there is an even number of variables that are -1 with the rest of them 1 or all of them 1.

It is equivalent to form a 7-digit number with a certain number of 1's & -1's.

$\begin{cases} \text{All of them 1} \to \text{ 1 solution} \\ \text{Exactly 2 of them -1,rest 1} \to \frac{7!}{5!2!} \\ \text{Exactly 4 of them -1,rest 1} \to \frac{7!}{4!3!} \\ \text{Exactly 6 of them -1,rest 1} \to \frac{7!}{6!1!}\end{cases}$

Number of solutions = $\binom{7}{0}+\binom{7}{1}+\binom{7}{2}+\binom{7}{3} = 2^{7-1} = 2^6=\boxed{64}$