Proof

[k3232x]

Q- Prove that when three consecutive even numbers are squared and the results are added, that the sum always has at least 3 divisors.

Thank you for your help.

 

[Denis]

All even numbers greater than 2 have 3 divisors:
1, 2 and itself.

Do you mean 3 divisors not including 1 and itself?

 

[galactus]

May I ask why you listed this in Geometry?.

Anyway, you could start by setting up a general expression for the sum of the

squares of 3 consecutive even integers:

$\displaystyle (2n)^{2}+(2n+2)^{2}+(2n+4)^2$

Expanding out and factoring gives:

$\displaystyle 4(3n^{2}+6n+5)$

As you can see, this has 2 and 4 as a divisor. Of course, the number itself is a divisor. That's three. Anymore are incidental.