$\forall n \in \mathbb{N}_{>5} \implies\exists (a,b)\in \mathbb Z^+:a^2+b^2\notin\mathbb P\;\wedge\;n=a+b$

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

  • $\begingroup$ There are $n/2$ (up to rounding) different choices for what $a^2 + b^2$ can be, ranging from $n^2/4$ (up to rounding) to $(n-1)^2 + 1$. Also, primes get rarer and rarer as you go further and further out on the number line. That seems to indicate that it's more and more probable that you'll be able to find some $a^2 + b^2$ which is composite. Of course, such a simple argument is no proof. $\endgroup$ – Arthur Aug 22 '17 at 6:19
  • $\begingroup$ I meant "ranging from $n^2/2$, up to rounding". Messed that up, apparently. $\endgroup$ – Arthur Aug 22 '17 at 6:26
  • $\begingroup$ Clearly it is just odds that you are looking at: If $n>5>2$ is even, then observe $(n-1)^2 + 1^2 = n^2 - 2n + 2$ is even and greater than $2$, hence non-prime. $\endgroup$ – Benjamin Dickman Aug 22 '17 at 6:52
  • 1
    $\begingroup$ If $n$ is composite, then the statement is easily provable. Indeed, let $n=x \cdot y$. Then you can take $(a,b)=(x, x(y-1))$, and $a^2+b^2$ is not prime since it's divisible by $x^2$. This reduces the study of this problem to primes larger than $5$. $\endgroup$ – Crostul Aug 22 '17 at 7:27

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 $(n-4)^2+4^2\equiv 0\pmod 5.$

  • $\begingroup$ For $n\not \equiv 0 \mod 5$ there exists $a$ such that $(2a-n)^2+n^2\equiv 0 \pmod 5,$ which is equivalent to $(n-a)^2+a^2\equiv 0\pmod 5.$ And $a\ne 0$ ( else $0\equiv (2a-n)^2+n^2\equiv 2n^2 \pmod 5$ ). $\endgroup$ – DanielWainfleet Aug 23 '17 at 5:24

We are looking for $n$ such that the quadratic polynomial $$\tag1 x^2+(n-x)^2\qquad(=2x^2-2nx+n^2)$$ is prime for surprisingly many integers $x$. It is well-known folklore that the polynomial $$\tag2 x^2+x+41$$ is prime for surprisingly many integers $x$. What is less (publicly) known is that this coincidence is owed to the fact that the discriminant of $(2)$, $-163$, has class number $1$ (which is rare). The discriminant of $(1)$ is $-n^2$ and therefore hardly of class number $1$.

To make this argument water-tight, one would have to elaborate more on the relation between class numbers and represented primes ...


This is not a solution, but it's too long for a comment.

As I wrote in the comment above, if $n$ is composite, then the statement is easily proved. Write $n=x \cdot y$ and pick $(a,b)=(x, xy-x)$. Since $$a^2+b^2=x^2(1+(y+1)^2)$$ is divisible by $x^2$, it's not a prime.

This reduces the study of this problem to primes $n \ge 7$.

This is equivalent to the following:

$$\exists a \in \left\{ 1, \dots , \frac{n-1}{2} \right\}: \ a^2+(n-a)^2 \mbox{ is composite}$$

Now, for all $a \in \left\{ 1, \dots , \frac{n-1}{2} \right\}$, call $$s=n-2a$$ Note that $s$ is odd and that $s \in \left\{ 1, 3, 5, \dots , n-2 \right\}$. Then $$\frac{n^2+s^2}{2} = \frac{2n^2-4an+4a^2}{2} = n^2-2an+2a^2 = a^2+(n-a)^2$$ which is composite for a suitable option for $s$.

Hence your conjecture is equivalent to the following:

For all primes $n \ge 7$ there exists an odd integer $s \in \{ 1, \dots , n-2\}$ such that $\frac{n^2+s^2}{2}$ is composite.


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