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

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The discussion centers on proving that the product of four consecutive natural numbers, f(n) = n(n+1)(n+2)(n+3), cannot be a perfect square. The argument presented shows that f(n) + 1 is a perfect square, specifically expressed as (n^2 + 3n + 1)^2. This leads to the conclusion that f(n) itself cannot be a perfect square due to the gap between consecutive square numbers. The gap between squares is emphasized as a crucial factor in understanding why f(n) cannot equal a square. The reasoning is acknowledged as correct, reinforcing the validity of the proof.
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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.
 
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Think about the size of the gap between consecutive square numbers.
 
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.
 
Just try a test case: what's the gap between x2 and (x-1)2?
 
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.
 
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