# The sum of two prime number is $999$. What is their product?

Q. The sum of two prime number is $999$. What is their product?

$997+2=999$ so all you would have to do is multiply $997$ and $2$

My question is what if there was another question like this but the prime numbers were in the middle and not so blatantly obvious like above. In other words, how would one approach this problem without the use of trial and error?

• If two primes are odd, then the sum is even. So, one of them is $2$. It related to this problem Twin prime – GAVD Sep 18 '15 at 4:37

In general, this problem is nontrivial. It's so nontrivial that we do not even know if it is possible to represent every even integer as a sum of two primes. This is one of the Goldbach conjectures.

In this case, there is a trick. The sum of your two primes is odd. This means that one of the primes is even, and the other is odd. But the only even prime is $2$.

HINT:

Any prime $>2$ is odd

Now odd $+$ odd $=$ even

For odd sum of two integers, the summands must be of opposite parity

In this case, $999$ is an odd number, so one of the two primes is $2$. That argument works for any odd number. In the case of an even number, I think the only approach is brute force, that is to check all decomposition until we find one.

The sum of two prime numbers cannot be an odd number unless one is even and the other is odd. And there is only one even prime number.

• PS: BTW $999=3\times3\times3\times37$. ${}\qquad{}$ – Michael Hardy Sep 18 '15 at 5:13

Just a comment to make clear what (usually) happens when an even number is used. Suppose the problem said, "The sum of two primes is 998, what is their product?". Then you could have:

7 + 991 = 998
31 + 967 = 998
61 + 937 = 998
79 + 919 = 998
139 + 859 = 998
211 + 787 = 998
229 + 769 = 998
241 + 757 = 998
271 + 727 = 998
307 + 691 = 998
337 + 661 = 998
367 + 631 = 998
379 + 619 = 998
397 + 601 = 998
421 + 577 = 998
457 + 541 = 998
499 + 499 = 998

All of these representations will (necessarily) lead to different products. So you could not solve the problem, in general, for an even sum. (It is expected but not proved that any even number greater than 12 will have more than one solution.)