Powers of two are not sums of squares of different integers greater than zero

Question

It seems that non-trivial sums of squares of different non-zero integers aren't powers of two. Tested for sums of $2,3,4$ terms.

Is there something known about this topic? Are there counter-examples?

Answer 1

Couldn't find a counter-example for up to $4$ terms as well (maybe someone can show why), but for more terms there are for example $$ 1^2+5^2+7^2+9^2+10^2=2^8\\ 2^2+3^2+5^2+7^2+13^2=2^8\\ 3^2+5^2+6^2+9^2+19^2=2^9\\ \vdots \\ 1^2+5^2+6^2+7^2+8^2+9^2=2^8\\ \vdots \\ $$

Answer 2

If all the squares are even, then there is a solution for a smaller power of two.
Odd squares are $1\pmod8$ so you need four of them to reach $4\pmod8$, and then an even square to reach $0\pmod8$. Sil's solutions each have four odd squares.

Answer 3

$$a^2+b^2=2^n$$

For $n\gt1$ right side is divisible by $4$, this forces $a=2a_1$ and $b=2b_1$.
$$(2a_1)^2+(2b_1)^2=2^n$$

Cancelling $4$ through out $${a_1}^2+{b_1}^2=2^{n-2}$$

Again if the right hand is divisible by $4$, this forces $a_1=2a_2$ and $b_1=2b_2$. Also cancelling $4$ through out $${a_2}^2+{b_2}^2=2^{n-4}$$ This can be continued till the right hand side becomes $2^1$ : $${a_i}^2 + {b_i}^2=2$$ This clearly has no solutions when $a_i\ne b_i$ or $a_i=0$ or $b_i=0$, so the original equation cannot have any nontrivial solutions when $a\ne b$.

Answer 4

Partial Answer

We prove there are no such sums for $p=2,3,4$. Let $$\sum_{i=1}^{p}a_i^2=2^k$$without loss of generality we may assume that at least one of the $a_i$s is odd (since otherwise we can divide the equality by two). Also we must have an even number of odd perfect squares leading to having at least two such. For the case of $p=2,3,4$ we cannot have exactly two odd perfect square since $$\sum_{i=1}^pa_i^2=8q_1+1+8q_2+1+4q_3=4q_4+2$$which can't be a power of two. Therefore the cases $p=2,3$ are eliminated.

For $p=4$ we are only left with the case where all the perfect squares are odd. Therefore$$\sum_{i=1}^4a_i^2=\sum_{i=1}^48q_i+1=8q_5+4$$which either can't be a power of two. Hence the proof is complete$\blacksquare$