Can you provide a counterexample to the claim given below ?
Inspired by Euler's conjecture about $8n+3$ (see page 4 of this note ) I have formulated the following claim :
For any nonnegative integer $n$ , $8n+6$ can be represented as a sum : $$8n+6 = a^2 +2p$$ , where $a$ is a nonnegative integer , and $p$ is a prime .
I have tested this claim for all $n$ up to $2 \cdot 10^6 $ .
I cannot prove the assertion, but I can prove that $p\geqslant 5$.
Proof: $$8n+6 = 2(4n+3)$$ and so is always even. Therefore, $a^2 + 2p$ must also be even, so $a^2$ is even and hence $a$ is also even. It follows, then, that for some $b\in\mathbb{Z^+}$, we have $a = 2b$. Thus, $$\begin{align} 2(4n+3) &= 4b^2 + 2p\tag*{$\big($Since $(2b)^2 = 4b^2\big)$} \\ &= 2(2b^2 + p).\end{align}$$ Therefore, $4n+3=2b^2+p$. Now we know that $2b^2$ can never be a square number itself (which is why $\sqrt{2}$ is irrational), so when $n = c^2$ for some $c\in\mathbb{Z}^+$, it follows that $4c^2\neq 2b^2$. And so, $p\neq 3$.
Now if $p=2$, then $4n+1=2b^2$ which is obviously absurd. Thus, $p > 2,3$ if (and only if!) $$p\geqslant 5.\tag*{$\bigcirc$}$$