Four Consecutive Integers Plus 1

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+1abcd + 1 be represented as (n−1)(n)(n+1)(n+2)+1=n4+2n3−n2−2n+1(n-1)(n)(n+1)(n+2) +1 = {n}^{4}+2{n}^{3}-{n}^{2}-2n + 1.

Factoring abcd+1abcd + 1 yields (n2+n−1)2{({n}^{2}+n-1)}^{2}, proving that it is a perfect square.

The average of adad and bcbc is

12[(n−1)(n+2)+n(n+1)]=12(2n2+2n−2)=n2+n−1\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+1abcd+1.

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

#Algebra #Average #PerfectSquare #Consecutivenumbers

Note by Steven Zheng
6 years, 10 months ago

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Comments

Solution

Let abcd+1abcd + 1 be represented as (n−1)(n)(n+1)(n+2)+1=n4+2n3−n2−2n+1(n-1)(n)(n+1)(n+2) +1 = {n}^{4}+2{n}^{3}-{n}^{2}-2n + 1.

Factoring abcd+1abcd + 1 yields (n2+n−1)2{({n}^{2}+n-1)}^{2}, proving that it is a perfect square.

The average of adad and bcbc is

12[(n−1)(n+2)+n(n+1)]=12(2n2+2n−2)=n2+n−1\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+1abcd+1.

Steven Zheng - 6 years, 10 months ago

It seems like a lot of work to expand (n−1)(n)(n+1)(n+2)+1(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⋯(1)((n-1)(n+2))((n)(n+1))+1\cdots (1)

Now let n2+n=xn^2+n=x.

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

Mursalin Habib - 6 years, 9 months ago

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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.

Steven Zheng - 6 years, 9 months ago
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