# How is this proof finished? I was told it is proven

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1. Sep 3, 2015

### AcousticBruce

Prove that the product of four consecutive natural numbers cannot be the square of an integer.

So let n be a natural number. So f(n) = n(n+1)(n+2)(n+3)

n --- 1 --- 2 --- 3 --- 4 --- 5 --- 10
f(n)-24--120-360-840-1080-17160

The conjecture I want to prove is F(n) + 1 is always a square.

n(n+1)(n+2)(n+3) = (n2+3n)(n2+3n+1)

so

[(n2+3n-1)+1)][(n2+3n+1)-1]+1

and because a2-b2 = (a+b)(a-b)

[(n2+3n+1)2-1)+1] = (n2+3n)2

So this proves that f(n) + 1 is in fact a square.

My question is how does this prove that f(n) is NOT a square? I mean it seems obvious, but I am trying to learn to prove things.

2. Sep 3, 2015

### TeethWhitener

Think about the size of the gap between consecutive square numbers.

3. Sep 3, 2015

### AcousticBruce

This is exactly what a friends said. The size gap. Ill think on this the rest of the day and see if I can find why that matters. Thanks.

4. Sep 3, 2015

### TeethWhitener

Just try a test case: what's the gap between x2 and (x-1)2?

5. Sep 6, 2015

### HomogenousCow

By the way, n(n+1)(n+2)(n+3) = (n2+3n)(n2+3n+2), so f(n)+1 = (n^2+3n+1)^2
The argument is correct, however.