Encouraging joy
Find four consecutive natural numbers. When you have found them, ask them to be quiet so you can multiply them. If the answer is a square number, jump up and down in joy.
Problem source: Wisconsin Mathematics Science & Engineering Talent Search.
February 17th, 2010 at 3:47 pm
Call the lowest of the numbers n. Then the product is
N = n(n+1)(n+2)(n+3)
= [n(n+3)][(n+1)(n+2)]
=(n²+3n)(n²+3n+2)
=(n²+3n+1)² – 1
i.e. N is 1 less than a square. N is a square only if
N = -1 or 0
so (n²+3n+1)² = 0 or 1
so n²+3n+1 = -1, 0 or 1
so n²+3n+2 = (n+1)(n+2) = 0, 1 or 2
(n+1)(n+2) = 0 entails n = -1 or -2.
One of the factors (n+1) and (n+2) is even (as n is an integer) so their product cannot be 1.
(n+1)(n+2) = 2 = (1)(2) = (-2)(-1) entails n=-3 or 0.
Negative numbers and zero are conventionally not regard as “natural numbers”. So, no joy.