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Let $p,q$ be distinct primes. Prove that the six integers $pq-2$, $pq+2$, $pq-4$, $pq+4$,$pq-6$, $pq+6$ cannot be all primes.

This is Exercise 5.60 in Chartrand's Mathematical Proofs. The claim is intriguing as for $p=3$ and $q=5$, the four integers $pq-2$, $pq+2$, $pq-4$, $pq+4$ are all primes. I haven't been able to explain why it stops working when $pq-6$ and $pq+6$ are added in the hypothesis.

I think I should look for a proof by contradiction. I don't know where to use the assumption that $p$ and $q$ are distinct primes.

I wonder if there is something peculiar about the prime gaps in this problem. I don't have any other clue.

Bill Dubuque
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Marlou marlou
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3 Answers3

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Let's assume that $pq+2$, $pq+4$ and $pq+6$ are all primes. I'm not considering the other 3 because the same logic will work for them too. One of these three numbers $pq+2$, $pq+3$ and $pq+4$ must be a multiple of 3. But since $pq+2$ and $pq+4$ already are primes therefore $pq+3$ is a multiple of 3. So, $pq+3=3k$ for some natural number $k$. which in turn implies that $pq+6$ which is just $(pq+3)+3 = 3(k+1)$ is also a multiple of $3$. Which contradicts our assumption that all three of these numbers are primes.

Doctorwho2311
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If you divide a prime number (other than $2$ or $3$) by 6, the remainder will be $1$ or $5$. If the remainder were $2$ or $4$ the number would be even, and if you got remainder $3$ the number would be divisible by $3$.

So if $p$ and $q$ are primes greater than $3$, then $pq$ will have remainder $1$ or $5$ when divided by $6$. If it is $1$ then $pq+2$ cannot be a prime (as it would have remainder 3, when divided by $6$). If it is $5$ then $pq+4$ cannot be a prime for the same reason.

Thus without loss of generality $p$ must be $2$ or $3$. However if $p=2$, then $pq+2$ will be even and greater than $2$. If $p=3$ then $pq+6$ will be divisible by $3$.

tkf
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Hint

Let $n=pq>6$.

Suppose $3$ does not divide $n-2$, then $n-2 \equiv 1,2 \pmod{3}$, then $n+2 \equiv 2,0 \pmod{3}$. But if $n+2 \equiv 2 \pmod{3}$, then $n+6 \equiv 0 \pmod{3}$.

So if $n-2$ is not divisible by $3$ then one of $n+2$ or $n+6$ has to be divisible by $3$.

Said differently, one of the numbers $n-2, n+2, n+6$ must be divisible by $3$.

Can you take it from here?

Anurag A
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  • It's implicit in your answer, but worth calling attention to, that $p,q$ being prime is a red herring. The fact is that for any $n$, $n\pm 2, n\pm 4, n\pm 6$ cannot all be primes. – Keith Backman Jul 25 '20 at 15:37