# Thread: Number theory, prime numbers, interesting problem

1. ## Number theory, prime numbers, interesting problem

Tell me does this problem have any solution? I cannot find any.... :blush: But I think this question is very interesting and have nice solution, help me:

numbers are given:

(1) ab + bc + ac

(2) a^3 + b^3 + c^3 - 2abc

Numbers a, b, c are prime numbers and we have to find what numbers a, b, c that (1) and (2) are divisible by (a + b + c).

So the question is: Find all (a, b, c) [a, b, c - prime numbers], that (1) and (2) are divisible by (a + b + c).

I am very sorry for my english. Can anybody help me?

2. Originally Posted by Ununuquantium
Tell me does this problem have any solution? I cannot find any.... :blush: But I think this question is very interesting and have nice solution, help me:

numbers are given:

(1) ab + bc + ac

(2) a^3 + b^3 + c^3 - 2abc

Numbers a, b, c are prime numbers and we have to find what numbers a, b, c that (1) and (2) are divisible by (a + b + c).

So the question is: Find all (a, b, c) [a, b, c - prime numbers], that (1) and (2) are divisible by (a + b + c).

I am very sorry for my english. Can anybody help me?
Let d = a + b + c. Then

$d(bc+ca+ab) = (b^2c+c^2a+a^2b) + (bc^2+ca^2+ab^2) + 3abc,$

$d^3 = (a^3+b^3+c^3) + 3(b^2c+c^2a+a^2b) + 3(bc^2+ca^2+ab^2) + 6abc,$

and therefore

$d^3 -3d(bc+ca+ab) = (a^3+b^3+c^3) -3abc = \{(a^3+b^3+c^3) -2abc\} - abc.$

If d divides $(a^3+b^3+c^3) -2abc$ then it follows that d divides abc. Since a, b and c are primes, and d is clearly greater than any of them, it follows that d must be equal to one of the products bc, ca, ab or abc. It is fairly easy to deduce from this that the only solution to the problem is $a=b=c=3.$

3. but for example a=2, b=3, c=5 ? (2*3*5) is divided by (2+3+5)