Maths Olympiad Prep

Library / /1 of 4

Number theory Difficulty 4.7 AIME Prove it Spain

Let aa, kk be positive integers and let nn be a nonnegative integer. Show that (ka2+1)2n+1(ka^2 + 1)^{2n+1} can be expressed as a sum of k+1k + 1 squares and (ka2+1)2n+2(ka^2 + 1)^{2n+2} can be expressed as a sum of (k+1)2(k + 1)^2 squares.

Solution

First, we have
(ka2+1)2n+1=(ka2+1)(ka2+1)2n=(a2+a2+⋯+a2+12)⏟k(ka2+1)2n=(a(ka2+1)n)2+(a(ka2+1)n)2+⋯+(a(ka2+1)n)2⏟k+((ka2+1)n)2⏟k (ka^2+1)^{2n+1} = (ka^2+1)(ka^2+1)^{2n} = \underbrace{(a^2+a^2+\dots+a^2+1^2)}_{k} (ka^2+1)^{2n} \\ = \underbrace{(a(ka^2+1)^n)^2 + (a(ka^2+1)^n)^2 + \dots + (a(ka^2+1)^n)^2}_{k} + \underbrace{((ka^2+1)^n)^2}_{k}
which is the sum of k+1k + 1 squares.

Likewise,
(ka2+1)2n+2=(ka2+1)2(ka2+1)2n=(k2a4+2ka2+1)(ka2+1)2n=(a2(ka2+1)n)2+(a2(ka2+1)n)2+⋯+(a2(ka2+1)n)2⏟k2+(a(ka2+1)n)2+(a(ka2+1)n)2+⋯+(a(ka2+1)n)2⏟2k+((ka2+1)n)2⏟k (ka^2 + 1)^{2n+2} = (ka^2 + 1)^2 (ka^2 + 1)^{2n} = (k^2a^4 + 2ka^2 + 1) (ka^2 + 1)^{2n} \\ = \underbrace{(a^2(ka^2 + 1)^n)^2 + (a^2(ka^2 + 1)^n)^2 + \dots + (a^2(ka^2 + 1)^n)^2}_{k^2} \\ + \underbrace{(a(ka^2 + 1)^n)^2 + (a(ka^2 + 1)^n)^2 + \dots + (a(ka^2 + 1)^n)^2}_{2k} + \underbrace{((ka^2 + 1)^n)^2}_{k}
is the sum of (k+1)2(k + 1)^2 squares, and we are done.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.