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Given four consecutive primes $p_1, p_2, p_3, p_4$ with $p_2-p_1=p_4-p_3$ and the division of $p_2*p_3$ by $p_1$ and $p_4$ giving the same remainder, is there a relationship between the value of these remainders and the pattern of gaps between the four primes?

For example, given 37, 41, 43, 47 the pattern of gaps is 4, 2 4 and the remainder is 24 for $\frac{41*43}{37}$ and for $\frac{41*943}{47}$, yet it is also 24 for 137, 139, 149,151 with a pattern of gaps 2,10,2.

The question is if you had the pattern of gaps could you know what the common remainders would be?

ewokx
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J. M. Bergot
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1 Answers1

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Let $m=\frac{p_2+p_3}2=\frac{p_1+p_4}2$ be a positive integer, and define $a$ and $b$ so that $(p_1,p_2,p_3,p_4)=(m-b,m-a,m+a,m+b)$. Then $$\frac{p_2p_3}{p_1}=\frac{m^2-a^2}{m-b}=m+b+\frac{b^2-a^2}{p_1}$$ and $$\frac{p_2p_3}{p_4}=\frac{m^2-a^2}{m+b}=m-b+\frac{b^2-a^2}{p_4};$$ so, as long as $b^2-a^2<p_1$, the product $p_2p_3$ has the same remainder $b^2-a^2$ modulo $p_1$ as modulo $p_4$. (Otherwise, at least for sufficiently large groups of four consecutive primes, the remainders cannot be the same, as this remainder would have to be equivalent modulo $p_1p_4$ to $b^2-a^2$, and thus be at most $b^2-a^2-p_1p_4=2b^2-a^2-m^2$, which is less than $0$ for sufficiently large primes.)

The gaps are $g_1=b-a$, $g_2=2a$, and $g_1=b-a$, and so the remainder is $$b^2-a^2=(b-a)(b+a)=g_1(g_1+g_2).$$

Carl Schildkraut
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  • How determine what sufficient large for primes means? Could this be done by simple experiment for large primes? – J. M. Bergot Jul 19 '22 at 02:41
  • Then the pair p2 and p3 would have to have a very large gap compared to gap p2-p1. Such a quartet of consecutive primes would not easily occur. Quite a search involved...any takers? – J. M. Bergot Jul 19 '22 at 03:12