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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 $ .

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    $\begingroup$ On page 240 of Mathematical Cranks, Underwood Dudley discusses the conjecture that every number of the form $4n+3$ can be written as the sum of a square and twice a prime, except 79. He opines the conjecture is false. $\endgroup$ – Gerry Myerson Apr 24 '18 at 12:50
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    $\begingroup$ Why does Dudley believe this is false? This strikes me as the sort of conjecture that's likely to be true except for a few small exceptions. $\endgroup$ – Michael Lugo Apr 24 '18 at 14:00
  • $\begingroup$ @GerryMyerson . I read "A Budget Of Trisectors" by Underwood Dudley. Is "Mathematical Cranks" a different version of it, or a separate book? $\endgroup$ – DanielWainfleet Apr 25 '18 at 2:12
  • $\begingroup$ @Michael, he doesn't give any reason. Note the use of the at-sign if you want to be sure someone see a comment. $\endgroup$ – Gerry Myerson Apr 25 '18 at 2:18
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    $\begingroup$ I can confirm that the conjecture is true for $8n + 6 \leq 10^9$. $\endgroup$ – Dylan Apr 26 '18 at 6:11
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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!) $$p\geqslant 5.\tag*{$\bigcirc$}$$

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    $\begingroup$ Your conjecture, in other words, is that $$4n+3=2b^2 + p\tag*{$\left(\begin{align}&\text{for non-negative integers $b$} \\ &\text{and $n$, and a prime $p\geqslant 5$.}\end{align}\right)$}$$ $\endgroup$ – Mr Pie May 13 '18 at 1:37

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