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For a prime $p$, define $\delta(p)$ to be the smallest offset $d$ from which $p$ differs from a square: $p = r^2 \pm d$, for $d,r \in \mathbb{N}$. For example, \begin{eqnarray} \delta(151) & = & +7 \;:\; 151 = 12^2 + 7 \\ \delta(191) & = & -5 \;:\; 191 = 14^2 - 5 \\ \delta(2748971) & = & +7 \;:\; 2748971= 1658^2 + 7 \end{eqnarray} For a particular $\delta=d$ value, define $\Delta(n,d)$ to be the number of primes $p$ at most $n$ with $\delta(p) = +d$, minus the number with $\delta(p) = -d$. In other words, $\Delta$ records the cumulative prevalence of $+d$ offsets over $-d$. For example, $\Delta(139,5)=-2$ because there are two more $-5$'s than $+5$'s up to $n=139$: $$ \delta(31)=-5 \;,\; \delta(41)=+5 \;,\; \delta(59)=-5\;,\; \delta(139) =-5 \;. $$ The figure below shows $\Delta(p,5)$ and $\Delta(p,7)$ out to the $200000$-th prime $2750159$. The offset $+7$ occurs $161$ times more than the offset $-7$, and the reverse occurs for $|\delta|=5$: $-5$ is more common than $+5$.

      PrimeSqOff57

Q. Is there a simple explanation for the different behaviors of offsets $5$ and $7$?

Obviously the question can be generalized to explaining the growth for any $|\delta|$.

I previously asked a version of this question on MSE, using somewhat different notation conventions and with less focused questions.

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  • $\begingroup$ The curve you plotted looks like a parabola. I don't know whether it can be explained though. $\endgroup$
    – Sylvain JULIEN
    Apr 27, 2017 at 11:12
  • $\begingroup$ The graph is a bit misleading, making the offsets seem small. In reality they are huge, once one keeps in mind that there are only about 1600 squares up to $2.7\times 10^6$ and and thus primes within $\pm 7$ (or $\pm 5$) of a square are only a smallish fraction of 3200 in number. I would be surprised if simple modular considerations don't explain the offsets. $\endgroup$
    – Yaakov Baruch
    Apr 27, 2017 at 11:34
  • $\begingroup$ And yes, in that case each curve would be close to $C \sqrt{n}/\log{n}$, so quite similar to a half parabola (and possibly with equal but opposite $C$ for the 2 cases). $\endgroup$
    – Yaakov Baruch
    Apr 27, 2017 at 11:45
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    $\begingroup$ This might have something to with with Chebyshev’s bias: note that $+7\equiv-5\equiv3\pmod4$. $\endgroup$
    – Emil Jeřábek
    Apr 27, 2017 at 12:04
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    $\begingroup$ The question title and text does not agree. Edit one of them. $\endgroup$
    – Stig Hemmer
    Apr 28, 2017 at 7:21

2 Answers 2

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Consider $n^2+7$ and $n^2-7$ modulo $3$. If these are to be prime they must be non-zero $\pmod 3$, and in the first case $n$ can be anything mod $3$, whereas in the second case $n$ must be $0 \pmod 3$. If you consider $n^2+5$ and $n^2-5$, you see that the pattern reverses. This is already a huge bias for one offset to be preferred over the other.

The Hardy-Littlewood conjectures make this precise. One expects that (for a number $k$ not minus a square) the number of primes of the form $n^2+k$ with $n\le N$ (say) is $$ \sim \frac 12 \prod_{p\ge 3} \Big(1 -\frac{(\frac{-k}{p})}{p-1} \Big) \frac{N}{\log N}, $$ where in the numerator of the product above is the Legendre symbol. The constants in front of the $N/\log N$ explain these biases.

You don't have to worry about a smaller offset than $\pm 5$ or $\pm 7$, since an application of the sieve shows that the numbers $n^2+a$ and $n^2+b$ (for fixed $a$, $b$) are both prime only $\le CN/(\log N)^2$ of the time for a constant $C$.

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    $\begingroup$ Lucia of course knows this, but I thought it was worth mentioning that this product is well-approximated by the inverse special value $L(1,\chi)^{-1}$ of the $L$-function associated to the unique Dirichlet character $\chi$ mod $4k$ such that $\chi(p) = \left( \frac{-k}{p} \right)$. Known bounds on this special value thus gives bounds on how large and small this constant can be. $\endgroup$
    – Will Sawin
    Apr 27, 2017 at 17:45
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Modulo 6 the squares are 0,1,4,3,4,1 and the squares+7 (or -5) can only be 1,2,5,4,5,2, of which 3/6 can at all be prime. The squares-7 (or +5) are 5,0,3,2,3,0 of which only 1/6 can be prime. Obviously this not a proof, but there is clearly no first order surprise in the observed offsets.

UPDATE I checked offsets modulo different numbers (such as (mod 2*3*5*7*11*13*17*...)) and I found that mod(3*7*11*19*23) we get 64638/25806 for +7/-7 and 21945/73899 for +5/-5, while primes 2, 5, 13, 17, 29 don't affect the ratio. I imagine that exploring more primes will further refine the ratios, with primes $\equiv 3$ (mod 4) affecting them and others not. WHAT IS GOING ON HERE?

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    $\begingroup$ I see now, thanks to Lucia's answer, that the ratios between the asymptotics there for, say, $k=\pm7$ are affected only by primes $p$ such that $$\Big(\frac{-7}{p}\Big)\neq \Big(\frac{7}{p}\Big)$$ which are the primes $\equiv 3$ (mod 4). $\endgroup$
    – Yaakov Baruch
    Apr 27, 2017 at 16:57

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