Ok. Let's start.
We want to proof that: The sum of the squares of 3 consecutive even numbers always has 3 divisors.


Lemma 1:
Every even number p can be written in the form p = 2k, with k∈ℕ.

Proof:
0 is even : 0 = 2k ; k=0
2 is even : 2 = 2k ; k=1
.
.
p is even : p = 2k ; k=p/2
Since p is even, p is a multiple of 2. Therefore p=2k', k'∈ℕ ⇒ k∈ℕ


Theorem:
If p is an even number, then p²+(p+2)²+(p+4)² has 3 divisors.
   p, p+2,p+4 are the 3 consecutive even numbers starting with p

Proof:
Let z = p²+(p+2)²+(p+4)²
I will write z|k if k is a divisor of z

We want to find {k1, k2, k3}∈ℕ³ so that:
z|k1, z|k2 and z|k3

It is known from number theory that every integer λ has always 2 trivial divisors: 1 and λ itself.

So z|1 and z|z. Therefore k1=1 and k2=z. We have found 2 divisors. We only need to find one more.

Let's get back to the expression.
z = p²+(p+2)²+(p+4)²

Since p is even, by our lemma 1 we can write p =2k, k∈ℕ
z = (2k)²+(2k+2)²+(2k+4)²
z = (2k)²+(2(k+1))²+(2(k+2))²
z = 2².k²+2².(k+1)²+2².(k+2)²
z = 2².(k²+(k+1)²+(k+2)²)

We can see that 2² is a factor of z. And since (k²+(k+1)²+(k+2)²) is an integer:

z divides 2², or z|2² = K3, the final divisor we needed.

So D|(z)={1, z, 4} are the trivial divisors of z. #D|(z) = 3. So z always has (at least) 3 divisors.

///c.q.d.

Last edited by kylekatarn (2005-11-21 11:47:04)

I did all that, but with p-2, p and p+2 for better cancellation.

I got it to 4(3p²+2) and then gave up because I forgot about the 1 and z bit, so I didn't post it.