Does there exist any $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of arithmetic square number.
Or
Can it be shown that
$$\begin{split}2^t&\ne \sum_{q=0}^u (n+qd)^2=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\end{split}$$
Where $n,u,d\in\mathbb{Z}_+$
Example: let $n=5,u=3,d=2$ so $5^2+7^2+9^2+11^2=276\ne 2^t$
$2^8= 1^2+5^2+7^2+9^2+10^2$ but $1,5,7,9,10 $ are not in arithmetic progression.
My incomplete attempt
Let $n,u,d\in\mathbb{Z}_+$
$$\begin{split}\sum_{q=0}^u (n+qd)^2 &=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\\ &=n^2(u+1)+\frac{(u+1)u}{2}(2nd+d^2)+\frac{(u+1)u(u-1)}{3}d^2 \end{split}$$
Let $$\begin{split}2^t &=\sum_{q=0}^u (n+qd)^2 \\ \implies 3\cdot 2^{t+1}&=6n^2(u+1)+3(u+1)u(2nd-d^2)+(u+1)u(u-1)2d^2 \\ &= (u+1)(6n^2+3u(2nd+d^2)+u(u-1)2d^2)\\ &(in\ case,\ u+1= 3) \\ \implies 2^t&= 3n^2+3(2nd+d^2)+2d^2\\ &= n^2+(n+1)^2+(n+2d)^2 \end{split}$$
Now we need to simplify for case, $6n^2+3u(2nd+d^2)+u(u-1)2d^2=3\cdot2^x$ and $u+1=2^y$ where $x+y=t+1$ but I'm stuck here. Thank you.
Related post:
Can a sum of consecutive $n$th powers ever equal a power of two?
Recent post
Let's focus on the case $n=d=1$.
$$\sum_{q=0}^u (n+qd)^2=\sum_{q=0}^u (1+q)^2=(u+1)+u(u+1)+\frac{u(u+1)(2u+1)}6$$ $$=(u+1)\frac{6+6u+u(2u+1)}6$$
Therefore, we have two cases: $\,u=2^k-1\,$ or $\,u=3\cdot 2^k-1$, for a certain $\,k$.
First case ($\,u=2^k-1\,$): $$\sum_{q=0}^u (1+q)^2=2^{k-1}\frac{6+6(2^k-1)+(2^k-1)(2(2^k-1)+1)}3=2^{k-1}\frac{6\cdot2^k+(2^k-1)(2^{k+1}-1)}3=2^{k-1}\frac{(3+2^{k+1})2^k+1}3$$
And so, necessarily,
$$(3+2^{k+1})2^k+1=3\cdot2^t$$ for a certain $\,t$. But this is impossible, because the two sides have different parity.
Second case ($\,u=3\cdot 2^k-1\,$): $$\sum_{q=0}^u (1+q)^2=2^{k-1}(6+6(3\cdot 2^k-1)+(3\cdot 2^k-1)(2(3\cdot 2^k-1)+1))=2^{k-1}(9\cdot 2^k(1+2^{k+1})+1)$$
So, in this case, necessarily
$$9\cdot 2^k(1+2^{k+1})+1=2^t$$ for a certain $\,t$. But, for the same reason as above, this is impossible.
I think that this argument can be extended to your general case.