# Prime Number Problem

Standard

Following is a problem about prime factorization of the sum of consecutive odd primes. (source: problem 80 from The Green Book of Mathematical Problems)

Prove that the sum of two consecutive odd primes is the product of at least three (possibly repeated) prime factors.

The first thing to observe is that sum of odd numbers is even, hence the sum of two consecutive odd primes will be divisible by 2. Let’s see factorization of some of the examples: $3 + 5 = 2\times 2 \times 2$ $5 + 7 = 2 \times 2\times 3$ $7+11 = 2 \times 3\times 3$ $11+13 = 2 \times 2 \times 2 \times 3$ $13+17 = 2 \times 3 \times 5$ $17+19 = 2\times 2\times 3 \times 3$ $19+23 = 42 = 2\times 3\times 7$ $23+29 = 52 = 2\times 2 \times 13$

Now let $p_n$ and $p_{n+1}$ be the consecutive odd primes, then from above observations we can conjecture that either $p_n+p_{n+1}$ is product of at least three distinct primes or $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ such that $k+\ell \geq 3$.

To prove our conjecture, let’s assume that $p_n+p_{n+1}$ is NOT a product of three (or more) distinct primes (otherwise we are done). Now we will have to show that if $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ then $k+\ell \geq 3$.

If $\ell = 0$ then we should have $k\geq 3$. This is true since $3+5=8$.

Now let $\ell > 0$. Since $k\geq 1$ (sum of odd numbers is even), we just need to show that $k=1, \ell=1$ is not possible. On the contrary, let’s assume that $k=1,\ell = 1$. Then $p_n+p_{n+1} = 2p$. By arithmetic mean property, we have $\displaystyle{p_n < \frac{p_n+p_{n+1}}{2}} = p

But, this contradicts the fact that $p_n,p_{n+1}$ are consecutive primes. Hence completing the proof of our conjecture.

This is a nice problem where we are equating the sum of prime numbers to product of prime numbers. Please let me know the flaws in my solution (if any) in the comments. Life is never fair. So I made my life a mathematical fair.

### 9 responses

1. OwnShadow on said:

I ‘discovered’ inadvertently that odd prime numbers are the only numbers that are the sum of two and only two consecutive integers. Do you happen to know how this description is defined in formal mathematical language?

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• OwnShadow on said:

I appreciate that, thank you.

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2. Andersays on said:

mine is less an attempt at flaws, than a question: Is 1 not considered prime? or os this solution only for all primes>1?

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• Andersays on said:

Thank you. I had always been taught that it was considered a prime

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• gaurish on said:

One way to justify that 1 is not a prime is that if it was prime then fundamental theorem of Arithmetic (unique factorization in terms of prime powers) won’t make any sense. I also had doubts regarding 1 being prime in school

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3. nkpithwa on said:

I went through it in detail. It seems quite an elegant solution. Similarly, you should also try the Scottish Book of problems (collections of problems posed and solved by Polish mathematicians like Stan Ulam, Stefan Banach…etc.) when Poland was a separate country before world war II. There is one more book “Red book of problems”…hopefully, all these books are there in your library…Once again, a very good attempt…:-)

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• gaurish on said:

Thank you. I am working on improving my solution writing skills. I have the red book, will also look at the Scottish book.

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