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Topic review (newest first)

2005-11-22 01:19:00

never give up : )

2005-11-21 17:52:18

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.

2005-11-21 17:03:54

We are not worthy ... we are not worthy ... (assembled masses bowing down)

But it seems like cheating to include 1 and z smile

2005-11-21 11:45:33

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.

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

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

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

We want to find {k1, k2, k3}∈ℕ 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 = p+(p+2)+(p+4)

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

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

z divides 2, or z|2 = 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.


2005-11-21 03:19:27

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.

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