Question by Đội bóng tia chớp FC - Discuss with MathYouLike

Đội bóng tia chớp FC

27/12/2017 at 20:49

For \(A=a^2+b^2+c^2\) where a , b are two consecutive natural numbers , c = ab . Prove that \(\sqrt{A}\) is an odd natural number.

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FA Liên Quân Garena 27/12/2017 at 20:56

\(A=a^2+b^2+c^2\)

\(=a^2+\left(a+1\right)^2+a^2\left(a+1\right)^2\)

\(=2a^2+2a+1+a^2\left(a+1\right)^2\)

\(=a^2\left(a+1\right)^2+2a\left(a+1\right)+1\)

\(=\left[a\left(a+1\right)+1\right]^2\)

We have : \(\sqrt{A}=a\left(a+1\right)+1\) is an odd number .
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