Thread: Always square

1x2x3x4 + 1 = 25 = 5^2
3x4x5x6 + 1 = 361 = 19^2
...


Prove that the product of 4 whole consecutive numbers plus one is always a perfect square.

bye

If by whole numbers you mean Z+,

For any whole number n,

n*(n+1)*(n+2)*(n+3)+1
= n^4+6*n^3+11*n^2+6*n+1
= (n^2+3*n+1)^2

(n^2+3*n+1)^2 is the square of (n^2+3*n+1), and (n^2+3*n+1) is an integer.

Originally Posted by paultwang

If by whole numbers you mean Z+,

As far as I can see (n^2+3n+1)^2 is a perfect square for all n \in Z, not only Z+...