Does this equation have a solution in the integers: $4^n= \dfrac{x^2+y^2+z^2+w^2}{2}$?

Question

Does this equation have a solution in the integers: $4^n= \dfrac{x^2+y^2+z^2+w^2}{2}$?

Obvious manipulation gives $x^2+y^2+z^2+w^2 = 2^{2n+1}\Rightarrow n\geq 0.$ I tried thinking of the case where only one of them is $1$ and the rest are zeros, but that would mean $n=-\dfrac{1}{2},$ which is impossible. After a little more experimentation, I'm pretty confident that there are no such solutions, but I'm stuck as to how to prove this. I think Fermat's Method of Infinite Descent may be useful here.

Answer 1

There are exactly $24$ solutions. You can answer this using modular forms (which gives a quantitative answer to the general Lagrange problem) but that is overkill, it is elementary to do directly, and see exactly what these $24$ solutions are.

Suppose that

$$a^2+b^2+c^2+d^2 = 2 \cdot 4^n.$$

Suppose for now that $n \ge 1$. Then the RHS is divisible by $8$. On the other hand, any square is either $0,1,4 \pmod 8$. By considering this mod $4$, either all terms are even or all terms are odd. But if all terms are odd, the LHS is $4 \pmod 8$. Hence all the terms are even, and then you get the equation

$$(a/2)^2 + (b/2)^2 + (c/2)^2 + (d/2)^2 = 2 \cdot 4^{n-1}.$$

By repeating this, you end up with the case of $n = 0$ where the arguments modulo $8$ do not occur. But clearly the solutions to $a^2+b^2+c^2+d^2=2$ are ju st the $24$ solutions given by permutations of

$$(\pm 1)^2 + (\pm 1)^2 + 0^2 + 0^2 = 1.$$

Hence there are $24$ solutions to the original equation given by permutations of

$$(\pm 2^n)^2 + (\pm 2^n)^2 + 0^2 + 0^2 = 2 \cdot 4^n.$$

Answer 2

By Lagrange's four-square theorem every natural number can be represented as the sum of four integer squares then also any

$$2\cdot 4^n$$

Answer 3

For the question as asked (i.e., are there any solutions, not how many?), two simple observations provide an easy answer. First of all, $4^n$ is a square. And second, for any square, we have

$$a^2={a^2+a^2\over2}={a^2+a^2+0^2+0^2\over2}$$

Remark: I'm not sure what kind of experimentation the OP did that led to the (incorrect) conclusion that there were no solutions, but one experiment to do in problems like this is to look at special cases. In particular, setting $n=0$ or $1$ gives the equations $2=x^2+y^2+z^2+w^2$ and $8=x^2+y^2+z^2+w^2$. It's easy to do an exhaustive analysis of the possible solutions here, since the only squares less than $2$ are $0$ and $1$ while the only ones less than $8$ are $0$, $1$, and $4$.