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This problem

Compute $$\sqrt{\sqrt{44\cdot 45 \cdot 46 \cdot 47+1}-44}$$

has a nice solution that relies on the identity

$$n(n+1)(n+2)(n+3) +1 = \left(n^2 + 3n + 1\right)^2$$

a word form of which is

The product of any four consecutive integers is one less than a perfect square.

Is that identity simply a mathematical coincidence - the coefficients just happen to be the same when you expand the left side as when you expand the right - or is there some reason for its truth?

I think this isn't a silly question, because identities like $$(n+1)^2 = n^2 + 2n +1$$ are very intuitive when you draw them:

enter image description here

(the green square represents $n^2$, the light orange rectangles are each $n$, and the dark orange square is $1$ by $1$)

Zubin Mukerjee
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1 Answers1

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The Key in here is that $$n(n+1)(n+2)(n+3) +1 $$ has a symmetry around $n+1.5$. 1.5 comes as the middle point between $0$ and $3$.

It is

$$(n+1.5-1.5)(n+1.5-0.5)(n+1.5+0.5)(n+1.5+1.5) +1 $$

Now, if we denote $n+1.5=m$ we get

$$(m-1.5)(m-0.5)(m+0.5)(m+1.5)+1$$ or $$(m^2-1.5^2)(m^2-0.5^2)+1=(m^2-2.25)(m^2-0.25)+1$$

Picking $x=m^2-2.25$ this becomes $x^2+2x+1$, which is your picture...

N. S.
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  • Thanks! The symmetry exists in larger products of consecutive integers, so is there a corresponding identity for something like $n(n+1)+1$ or $n(n+1)(n+2)(n+3)(n+4)(n+5)+1$? – Zubin Mukerjee Oct 09 '14 at 23:33
  • @ZubinMukerjee The symmetry makes things nicer, but doesn't always lead to a good identity. What does the symmetry really does, it expresses the identity in terms of $m^2$, which means that the Polynomial of degree $2n$ actually becomes a Polynomial of degree $n$ in $m^2$. This way it gets nicer, but most of the times you don't get a perfect square. – N. S. Oct 09 '14 at 23:44