Let $a$ and $b$ be positive integers and $K=\sqrt{\frac{a^2+b^2}2}$, $A=\frac{a+b}2$. If $\frac KA$ is a positive integer, prove that $a=b$.
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Tags: power mean, Divisibility, algebra, number theory
MathLuis
11.04.2021 05:40
If $\frac{K}{A}$ is an integer then its square must be an integer so lets start: $\frac{\frac{a^2+b^2}{2}}{\frac{a^2+2ab+b^2}{4}}=\frac{2(a^2+b^2)}{a^2+2ab+b^2}$ now since $a^2+b^2 < (a+b)^2$ for all positive integers then $2(a^2+b^2)=c(a+b)^2$ for some $c$ on possitive integers (note that $c<2$) and then $2a^2+2b^2=a^2+2ab+b^2$ for some $a,b$. Simplifying you get: $(a-b)^2=0$ and then $a=b$ thus we are done
jasperE3
11.04.2021 05:45
MathLuis wrote:
$\frac KA=\frac{2a^2+2b^2}{a^2+2ab+b^2}$
$<\frac{2a^2+4ab+b^2}{a^2+2ab+b^2}=2$
$\Rightarrow\frac{2a^2+2b^2}{a^2+2ab+b^2}=1\Rightarrow a=b$
sorry for stealing this from sqing
MathLuis
11.04.2021 05:47
jasperE3 wrote:
MathLuis wrote:
$\frac KA=\frac{2a^2+2b^2}{a^2+2ab+b^2}$
$<\frac{2a^2+4ab+b^2}{a^2+2ab+b^2}=2$
$\Rightarrow\frac{2a^2+2b^2}{a^2+2ab+b^2}=1\Rightarrow a=b$
sorry for stealing this from sqing No worries, its the same think but more redacted an reduced