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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!

Peter
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Lehs
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  • There where two conjectures and the remaining is: $n=p+a^2+b^2$. – Lehs Aug 05 '18 at 14:46
  • "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? – Keith Backman Aug 05 '18 at 17:33
  • @KeithBackman This is Lagrange's four square theorem, and four is smallest number. There is also Lagrange's three square theorem which gives necessary and sufficient conditions for $n$ being sum of three squares. Basically it implies whenever $n$ is of form $4^a(8b+7)$, then you cannot express it as a sum of three squares (but you can still express it as four squares). – Sil Aug 05 '18 at 17:49
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    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. – Sil Aug 05 '18 at 17:52
  • @Sil Any natural number is the sum of no more than four squares, but the question implied there might be something special about using exactly four (not fewer) squares (maybe I read too much into that statement?). No four squares add up to $3, 5, 6, 9, 11,..$ unless one or more of the squares is $0^2$. These numbers are the sums fewer than four squares, of course. I was asking, Is there a smallest natural number such that all numbers larger than it can be decomposed into exactly four squares not including $0^2$ (whether or not fewer squares are sufficient)? – Keith Backman Aug 05 '18 at 19:49
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    @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). – Sil Aug 05 '18 at 20:10
  • @Lehs $(1)$ Just to clarify : Is $0$ an allowed square ? $(2)$ What might help : A positive integer can be written as the sum of at most two squares, if and only if every prime factor of the form $4k+3$ occurs with an even exponent in the prime factorization. $(3)$ The given claim should be much weaker than Goldbach's conjecture, so I am optimistic that a proof is possible, however I have no clue yet how. – Peter Aug 06 '18 at 11:03
  • An even stronger statement seems to hold : Every odd number $n\ge 61$ can be written as $n=p+2q$ with odd primes $p$ and $q$ and $q\equiv 1\mod 4$. This conjecture holds for $n<10^7$ and my program currently verifies the range $[10^7,10^8]$. This stronger conjecture implies the given statement for the odd integers greater than $59$. – Peter Aug 06 '18 at 12:21
  • Similar, it seems that every even number $n\ge 64$ can be wrtiiten as $n=p+q$ with odd primes $p$ and $q$ and $q\equiv 1\mod 4$. This is true for $n\le 10^7$ and implies the given conjecture for even integers greater than $62$. – Peter Aug 06 '18 at 12:29
  • Both my stronger conjectures hold upto $n\le 10^8$ , hence the given conjecture is true for $n\le 10^8$ – Peter Aug 06 '18 at 13:18
  • @Peter: thanks for your interesting comments! In my conjecture the squares can also be zero. I hope it's weak enough to be able to prove. – Lehs Aug 06 '18 at 13:59
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    @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 ... – Peter Aug 07 '18 at 16:30

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