1. ## Prove this proposition

Prove :

2. here's a solution:

Suppose $\frac{a^2+b^2}{a^2-b^2}$ is an integer. Then we have

$\frac{a^2+b^2}{a^2-b^2} = \frac{a^2-b^2}{a^2-b^2}+\frac{2b^2}{a^2-b^2} = 1+\frac{2b^2}{a^2-b^2} \in \mathbb{N}\Leftrightarrow\frac{2b^2}{a^2-b^2} \in \mathbb{N}$

$\Leftrightarrow 2b^2 = n(a^2-b^2), n\in\mathbb{N} \Leftrightarrow (n+2)b^2 = na^2 \Leftrightarrow$

Set $p=\frac{a}{gcd(a,\ b)}$ and $q=\frac{b}{gcd(a,\ b)},$ where gcd(a, b) is the greatest common divisor of a and b. Note that both p and q are integers but that they don't have any common prime factor.

$(n+2)b^2 = na^2\Leftrightarrow(n+2)q^2=np^2\Leftrightarrow\fr ac{n+2}{p^2}=\frac{n}{q^2}=\frac{j}{k},\ j,\ k \in\mathbb{N},\ gcd(j,\ k) = 1$

Every prime factor in k also has to be in $p^2$ and $q^2$, but since $gcd(p,\ q)=1$, $p^2$ and $q^2$ share no prime factors. So $k = 1$

$\therefore\ n+2=jp^2\text{ and }n=jq^2$

$p^2=q^2+\frac{2}{j} < q^2+2q+1=(q+1)^2$

$a > b\Leftrightarrow p > q$

$\therefore\ q < p < q+1$

But there are no integers in the interval (q, q+1).

So $\frac{a^2+b^2}{a^2-b^2}$ cannot be an integer.

3. ## Simpler

Suppose $\frac{a^2+b^2}{a^2-b^2}=n\in\mathbb{N}$. (Start by noticing $n\neq1$ else $b=0$.) Rearranging, $\frac{a^2}{b^2}=\frac{n+1}{n-1}=\frac{n^2-1}{(n-1)^2}$. So $n^2-1=\frac{a^2}{b^2}(n-1)^2$. Since $n\neq1$, the RHS is a perfect nonzero square. The LHS is one less than a perfect square. No two squares differ by only one in the naturals. QED

4. Nice.