Proof about product of 4 integers

[cragar]

Homework Statement

Prove that the product of four consecutive integers is always one less than a perfect square.

The Attempt at a Solution

I tried looking at the product $(n-1)(n)(n+1)(n+2)=x^2-1$
but i couldn't seem to get anything useful out of it. I added one to both sides .
I tried to see if i could some how show that the product of the four integers plus 1 had an odd number of divisors but I couldn't see a way to do it. I did notice that these numbers are never divisible by 2,3,or 4, and a lot of the time they produce a prime, all though I didn't check very many.
And when it wasn't a prime it was divisible by 5. I was trying to think of a way to do this with Pythagorean triples but I am not sure.

 

[AlephZero]

I tried looking at the product $(n-1)(n)(n+1)(n+2)=x^2-1$
but i couldn't seem to get anything useful out of it.

Look harder . Don't overthink the problem with the other ideas you tried. This doesn't depend on any theorems in number theory!

It's hard to give a hint without telling you the answer, but think about what you can you do with $x^2-1$.

 

[cragar]

are you talking about factoring it into (x+1)(x-1) I not sure how that helps I did look at that. we do know that x needs to be odd. so x=2y+1 so (2y+1+1)(2y+1-1)=(2y+2)(2y)=4y(y+1) I still don't see how it will work. I guess Ill keep trying stuff. thanks for your post.

 

[AlephZero]

are you talking about factoring it into (x+1)(x-1) I not sure how that helps

Don't bother about x being odd or even. Just factor the left hand side into the same form.

 

[cragar]

So we have (n^2-1)(n)(n+2)=x^2-1 I just don't see what this does, I can see how we have similar things on both sides. but it seems like we still have 2 extra terms on the left side.

 

[AlephZero]

Try taking different pairs of factors from $(n-1)(n)(n+1)(n+2)$ to get $(an^2 + bn + c -1)(an^2 + bn + c + 1)$.

Or, factorize $(n-1)(n)(n+1)(n+2) + 1$ into a perfect square, like $(an^2 + bn + c)^2$. It's not hard to guess the values of $a$ and $c$.

 

[cragar]

ok thanks for your help, I see it now