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=n4+2n3−n2−2n+1.
Factoring abcd+1 yields (n2+n−1)2, proving that it is a perfect square.
The average of ad and bc is
21[(n−1)(n+2)+n(n+1)]=21(2n2+2n−2)=n2+n−1
which is the square-root of abcd+1.
Check out my other notes at Proof, Disproof, and Derivation
#Algebra
#Average
#PerfectSquare
#Consecutivenumbers
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Solution
Let abcd+1 be represented as (n−1)(n)(n+1)(n+2)+1=n4+2n3−n2−2n+1.
Factoring abcd+1 yields (n2+n−1)2, proving that it is a perfect square.
The average of ad and bc is
21[(n−1)(n+2)+n(n+1)]=21(2n2+2n−2)=n2+n−1
which is the square-root of abcd+1.
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⋯(1)
Now let n2+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.