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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 \(abcd+1\).

Log in to reply

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 \((x-1)^2\).

Log in to reply

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.

Log in to reply