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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TopNewestSolutionLet $abcd + 1$ 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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It 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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