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High School Math Grades 9-12, Ages 13-18

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#1 h Sep 26, 2017, 10:56 AM • 1 Y

Y by Adventure10

Problem C:2534,page 356, GMB 9/2002
Let $a,b,n$ be positive integers such that
$\frac{a+n}{b+n}=n\cdot \frac{a}{b}$ . Prove that $\frac{b}{a}$ is a perfect square.
(Author Petru Raducanu, Iasi)

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#2 h Sep 26, 2017, 3:11 PM • 1 Y

Y by Adventure10

Answer: $\frac{b}{a}=n^{2}$ ;
more precisely: $a=1\Rightarrow b=n^{2},a=2\Rightarrow n=1 andb=2,a> 2\Rightarrow n=1 and b=a$ .
\center Solution
$\centering$ $\frac{a+n}{b+n}= \frac{n\cdot a}{b}\Leftrightarrow$
$ab+nb=nab+n^{2}a$ $(1)$
$\Rightarrow n^{2}\cdot \frac{a}{b}=a-n(a-1).$
\raggedright So we have $a-n(a-1)>0$ $(2)$
If in (2) we have $a=1$ then from (1) we get $b=n^{2}$ so $\frac{b}{a}=n^{2}$ .
If in (2) we have $a=2$ , meaning $2-n>0$ , then necessary $n=1$ , and from (1) we get $b=a$ so again $\frac{a}{b}=1=n^{2}$
If $a>2$ then from (2)result
$n<\frac{a}{a-1}$ ; $(3)$
but $a>2\Leftrightarrow 2a>a+2\Leftrightarrow 2a-2>a\Leftrightarrow \frac{a}{a-2}<2$ ,and then (3) involve $n<2$ .So $n=1$ and from (1) we get $b=a$ ,so again $\frac{a}{b}=1=n^{2}$ .

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