<s>[INST] $\forall n \in \mathbb{N}_{>5}\implies\exists (a,b)\in \mathbb Z^+:a^2+b^2\notin\mathbb P\;\wedge\;n=a+b$ Conjecture:
$\forall n \in \mathbb{N}_{>5} \implies\exists (a,b)\in \mathbb Z^+:a^2+b^2\notin\mathbb P\;\wedge\;n=a+b$
Tested $\forall n\leq 100,000$. Small exceptions: {1,2,3,5}.
I would like to see proofs or counterexamples.
The conjecture seems to be related to:
Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$
Any odd number is of form $a+b$ where $a^2+b^2$ is prime
 [/INST] If $n=5m$ with $m>1$ then $n=4m+m$ and $(4m)^2+m^2=17m^2$ is composite because $m^2>1.$ 
If $n\equiv 4 \pmod 5$ then $(n-1)^2+1^2>(n-1)^2\geq 3^2>5$ and $(n-1)^2+1^2 \equiv 0\pmod 5.$ 
If  $5<n\equiv 1\pmod 5$ then $(n-2)^2+2^2>(n-2)^2\geq 4^2>5$ and $(n-2)^2+2^2\equiv 0\pmod 5.$ 
If $5<n\equiv 3\pmod 5$ then $(n-2)^2+2^2>(n-2)^2\geq 6^2>5$ and $(n-2)^2+2^2\equiv 0 \pmod 5.$ 
If $5<n\equiv 2 \pmod 5$ then $(n-4)^2+4^2>(n-4)^2\geq 3^2>5$ and $(n-4)^2+4^2\equiv 0\pmod 5.$
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