Positive integers have the form $a^2+b^2+c^2+2^d$

Question

For any positive integer $n$ there seems to be non-negative integers $a,b,c,d$ such that

$$n=a^2+b^2+c^2+2^d.$$

Due to Legendre's three-square theorem a natural number can be represented as the sum of three squares of integers

$$m=a^2+b^2+c^2$$

if and only if $m$ is not of the form $m=4^a(8b+7)$, for integers $a$ and $b$. If the conjecture is true, then for all $n\geq 1$ it must exist a $k$ such that $n-2^k$ is not of the form $m=4^a(8b+7)$.

I want help to prove the conjecture or to find a counter-example.

Answer 1

We claim that $n$ can be written as $a^2+b^2+c^2+x$ where $x\in\{1,2,4\}$. Indeed, otherwise, we must have that

$$n-1,n-2,n-4$$

are each of the form $4^a(8b+7)$. However, all such numbers are either $\equiv 0$ or $\equiv 3\bmod 4$, while $n-1,n-2,n-4$ are each members of different residues $\bmod 4$. So, it only remains to prove this in the case for which these are not all positive integers, namely $n\leq 4$. We have

$$0^2+0^2+0^2+2^0=1$$

$$0^2+0^2+0^2+2^1=2$$

$$0^2+0^2+1^2+2^1=3$$

$$0^2+0^2+0^2+2^2=4.$$