Let \(a, b, c, d\) be four consecutive integers. Prove that \(abcd + 1\) is a perfect square. Furthermore, prove that the the square root of \(abcd + 1\) is equal to the average of \(ad\) and \(bc\).

**Solution**

Let \(abcd + 1\) be represented as \((n-1)(n)(n+1)(n+2) +1 = {n}^{4}+2{n}^{3}-{n}^{2}-2n + 1\).

Factoring \(abcd + 1\) yields \({({n}^{2}+n-1)}^{2}\), proving that it is a perfect square.

The average of \(ad\) and \(bc\) is

\[\frac{1}{2}\left[(n-1)(n+2) + n(n+1)\right] = \frac{1}{2}(2{n}^{2} +2n -2 )= {n}^{2}+n-1\]

which is the square-root of \(abcd+1\).

Check out my other notes at Proof, Disproof, and Derivation

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## Comments

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TopNewestIt seems like a lot of work to expand \((n-1)(n)(n+1)(n+2)+1\) and factor it. Here's a cool trick.

Rearrange it like this,

\(((n-1)(n+2))((n)(n+1))+1\cdots (1)\)

Now let \(n^2+n=x\).

\((1)\) simplifies to \((x-2)(x)+1\) which is equal to \((x-1)^2\).

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Nice! I solved this a long, long, time ago and I copied this from my notebook. Back then, I was just learning algebra, and brute forced these things.

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SolutionLet \(abcd + 1\) be represented as \((n-1)(n)(n+1)(n+2) +1 = {n}^{4}+2{n}^{3}-{n}^{2}-2n + 1\).

Factoring \(abcd + 1\) yields \({({n}^{2}+n-1)}^{2}\), proving that it is a perfect square.

The average of \(ad\) and \(bc\) is

\[\frac{1}{2}\left[(n-1)(n+2) + n(n+1)\right] = \frac{1}{2}(2{n}^{2} +2n -2 )= {n}^{2}+n-1\]

which is the square-root of \(abcd+1\).

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