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I have found that the primes which divide $(2N)^2+1$ is of the form $4m+1$.

I wonder if it is true that the primes which divide $(2N)^2+3$ is not of the form $4m+1$.

Kim Hollesen
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  • By "specific form" for the "primes which divide $(2N)^2+1$", do you mean that they are all of the form $4m + 1$? If so, then note that $(2(3))^2+3=39=3(13)$, where $13=4(3)+1$. – John Omielan Oct 03 '23 at 06:06
  • Also, we have that $(2(9))^2 + 1 = 325 = 13(25)$. Thus, $13$ is a prime which divides numbers of both of the forms that you specified. Please add a few details, in particular what you mean by "specific form", and why you think that primes which divide $(2N)^2+3$ would have a "distinct" form (in addition, please also specify what you mean by "distinct") to your question by using the Edit button. – John Omielan Oct 03 '23 at 06:16
  • This might go without saying, but there certainly aren't any odd primes that divide both numbers: if an integer $m$ (perhaps prime, or perhaps not) divides both $n + 3$ and $n + 1$ (here $n$ is any integer, perhaps of the form $(2N)^2$, or perhaps not), then $m$ must divide the difference $(n+3)-(n+1) = 2$. – leslie townes Oct 03 '23 at 08:47
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    John OMielan. Yes, I was hoping that the primes which divides $N^2+1$ is always distinct from the primes which divide $N^2+3$. Your example shows that it is not the case. Since $5^2+1= 213$ and $6^2+3=133$ – Kim Hollesen Oct 03 '23 at 14:23

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No, since $37 = (2*3)^2+1$ and $(2*8)^2+3 = 259 = 37*7$

PhyX_Meow
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