Nice! If $a=b$, everything is clear. If $a<b$, then \[\frac ab<\frac{a+d}{b+d}\iff\frac{b-a}b>\frac{b-a}{b+d}\iff d>0\]If $a>b$, then \[\frac ab>\frac{a+d}{b+d}\iff\frac{a-b}b>\frac{a-b}{b+d}\iff d>0\] Thus, we can note that $\frac ab=\frac{a+d}{b+d}$ only if $d=0$. Thus, as \[\frac ab=\frac{a^2+ab}{b^2+ab}\]we must have $n^2=ab$.

naman12 wrote: Nice! If $a=b$, everything is clear. If $a<b$, then \[\frac ab<\frac{a+d}{b+d}\iff\frac{b-a}b>\frac{b-a}{b+d}\iff d>0\]If $a>b$, then \[\frac ab>\frac{a+d}{b+d}\iff\frac{a-b}b>\frac{a-b}{b+d}\iff d>0\] Thus, we can note that $\frac ab=\frac{a+d}{b+d}$ only if $d=0$. Thus, as \[\frac ab=\frac{a^2+ab}{b^2+ab}\]we must have $n^2=ab$. This 'solution' doesn't make sense at all. You need to put a bit more explanation at the end.

We can just cross multiply and get $ab = n^2$ (holds true even when $a=b$) @naman12 why did you do case work?

Lemma: If $\frac{x}{y} = \frac{z}{t}$ then for all non zero real numbers $k$ we have $\frac{x}{y} = \frac{z-kx}{t-ky}$ Proof: Trivial Apply lemma: $$ \frac{a}{b} = \frac{a^2+n^2}{b^2+n^2} = \frac{n^2}{b^2+n^2 -ab} = \frac{a^2+n^2 - ab}{n^2}$$Now we just expand: \begin{align*} n^4 = (a^2+n^2-ab)(b^2+n^2-ab) \\ n^4 = a^2b^2 n^2a^2 -a^3b +n^2b^2 - ab^3 +n^4 -2n^2ab + a^2b^2 \\ n^2(a-b)^2 = ab(a-b)^2 \end{align*}If $a=b$, we are done, otherwise we have $n^2=ab$ and the conclusion follows.

dangerousliri wrote: This 'solution' doesn't make sense at all. You need to put a bit more explanation at the end. Which part doesn't make sense? I was clearly extremely tired last night, so pardon if I messed up. MrOreoJuice wrote: We can just cross multiply and get $ab = n^2$ (holds true even when $a=b$) @naman12 why did you do case work? My first instinct was that lemma, but I forgot the "good" proof, so I did casework.

Just factor this as:$(ab-n^2)(a-b)=0$ then: $a=b$ gives $ab=a^2$ OR $ab=n^2$

$\color{blue} \boxed{\textbf{SOLUTION}}$ $\color{red} \textbf{Case 1: a,b}$ are different $\frac{a}{b}=\frac{a^2+n^2}{b^2+n^2}$ $\implies \frac{a-b}{b}=\frac{a^2+n^2-b^2-n^2}{b^2+n^2}$ $\implies \frac{1}{b}=\frac{a+b}{b^2+n^2}$ $\implies b^2+n^2=b^2+ab$ $\implies n=\sqrt {ab} \in \mathbb Z$ $\color{red} \textbf{Case 1: a=b}$ If $a=b$ then, $\sqrt {ab} = \sqrt {a^2} = a \in \mathbb Z \blacksquare$

kfas wrote: Positive integers $a$, $b$ an $n$ satisfy \[ \frac{a}{b}=\frac{a^2+n^2}{b^2+n^2}. \]Prove that $\sqrt{ab}$ is an integer. It is written $ab(b-a)=n^2(b-a)$. If $a=b$, trivial. Else, $\sqrt{ab}=n$.