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I'm not sure if this conjecture is less hard than Goldbachs conjecture:

any integer greater than $2$ is the sum of an odd prime and two squares of integers.

Facts as:

  • Every prime of the form $4n+1$ is the sum of two squares.

  • Every natural number is the sum of four squares

may or may not be helpful.

I've tested the conjecture for all integers less than $10^6$.

Even if the conjecture maybe is to hard to prove I would like to see ideas and heuristics about it. Or counter-examples!

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  • $\begingroup$ There where two conjectures and the remaining is: $n=p+a^2+b^2$. $\endgroup$ – Lehs Aug 5 '18 at 14:46
  • $\begingroup$ "Any natural number is the sum of four squares": For some natural numbers, this can only be accomplished by allowing $0$ to be one (or more) of the numbers in the sum. But once you allow that, you may as well say that any natural number is the sum of $434,738$ or any other arbitrary number of squares. Is there a lower limit above which four non-zero squares make the statement true? $\endgroup$ – Keith Backman Aug 5 '18 at 17:33
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    $\begingroup$ By the way, $n$ can be expressed as a sum of two squares iff all primes $p\mid n$ of form $p=4k+3$ divide $n$ in even power. $\endgroup$ – Sil Aug 5 '18 at 17:52
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    $\begingroup$ @KeithBackman If I understand what you are asking, that is not true for four squares, because $2^{2r+1}$ can be expressed in exactly one way $2^{2r+1}=0^2+0^2+(2^r)^2+(2^r)^2$ (can be proven by induction on $r$). But for five squares the result holds for $n>33$ (i.e. every such $n$ can be expressed as sum of five non-zero squares). $\endgroup$ – Sil Aug 5 '18 at 20:10
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    $\begingroup$ @Lehs I have no proof yet, but I have a somewhat stronger statement that seems to be true : Every positive integer $n\ge 3$ can be written as a sum of an odd prime and a non-negative integer that has no prime factor of the form $4k+3$. This is true upto $n\le 10^8$ and heuristically , it should always be true. All we need is a prime $p\le n$ such that $n-p$ is not divisible by $3,7,11,19,23,31,43,\cdots$. Maybe , someone can work that out ... $\endgroup$ – Peter Aug 7 '18 at 16:30

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