This is an answer to the second question, "what is the simplest direct proof?"

Let m be an integer satisfyng $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}=m$. $a,b,c,d \in \mathbb{Z}$, gcd(a,b)=1, gcd(c,d)=1.

$\frac{a^{2}d^{2}+b^{2}c^{2}}{b^{2}d^{2}}=m$

$a^{2}d^{2}+b^{2}c^{2}=mb^{2}d^{2}$

$b^{2}(md^{2}-c^{2})=a^{2}d^{2}$.

As our assumtion $b^{2} \not | a^{2}$ so $b^{2} | d^{2}$, $d^{2}=kb^{2}$. Then $\frac{1}{b^{2}}(a^{2}+\frac{c^{2}}{k})=m$.

$\frac{c^{2}}{k}=mb^{2}-a^{2}$ which is an integer. Thus we have $k=1$ as gcd(c,d)=1.

We have $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{b^{2}}=m$, $a^{2}+c^{2}=mb^{2}={p_{1}}^{\alpha_{1}}...{p_{k}^{\alpha_{k}}}$ [Unique prime factorization].

If $a^{2}+c^{2}=wp$, p is prime, then we can find integers x, y such that $x^{2}+y^{2}=p$.

Thus for every $p_{1}$, $p_{2}$,...,$p_{k}$, we can find $a_{1}, a_{2},...,a_{k}$ and $c_{1}, c_{2},...,c_{k}$ such that

${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$

${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$

$\vdots$

${a_{k}}^{2}+{c_{k}}^{2}=p_{k}$

Now if $m=p_{1}p{2}$, ${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$, ${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$.

$({a_{1}}^{2}+{c_{1}}^{2})({a_{2}}^{2}+{c_{2}}^{2})$ = ${(a_{1}a_{2}+c_{1}c_{2})}^{2}+ {(a_{1}c_{2}-c_{1}a_{2})}^{2}=m$.

If $m={p_{1}}^{2}$, then $m={({a_{1}}^{2}+{c_{1}}^{2})}{({a_{1}}^{2}+{c_{1}}^{2})}={({a_{1}}^{2}+{c_{1}}^{2})}^{2}+0$. We can continue in this way to show the claim.

This is an answer to the second question, "what is the simplest direct proof?"

Let m be an integer satisfyng $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}=m$. $a,b,c,d \in \mathbb{Z}$, gcd(a,b)=1, gcd(c,d)=1.

$\frac{a^{2}d^{2}+b^{2}c^{2}}{b^{2}d^{2}}=m$

$a^{2}d^{2}+b^{2}c^{2}=mb^{2}d^{2}$

$b^{2}(md^{2}-c^{2})=a^{2}d^{2}$.

As our assumtion $b^{2} \not | a^{2}$ so $b^{2} | d^{2}$, $d^{2}=kb^{2}$. Then $\frac{1}{b^{2}}(a^{2}+\frac{c^{2}}{k})=m$.

$\frac{c^{2}}{k}=mb^{2}-a^{2}$ which is an integer. Thus we have $k=1$ as gcd(c,d)=1.

We have $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{b^{2}}=m$, $a^{2}+c^{2}=mb^{2}={p_{1}}^{\alpha_{1}}...{p_{k}^{\alpha_{k}}}$ [Unique prime factorization].

If $a^{2}+c^{2}=wp$, p is prime and gcd(w,p)=1, then we can find integers x, y such that $x^{2}+y^{2}=p$.

Thus for every $p_{1}$, $p_{2}$,...,$p_{k}$, we can find $a_{1}, a_{2},...,a_{k}$ and $c_{1}, c_{2},...,c_{k}$ such that

${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$

${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$

$\vdots$

${a_{k}}^{2}+{c_{k}}^{2}=p_{k}$

Now if $m=p_{1}p_{2}$, ${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$, ${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$.

$({a_{1}}^{2}+{c_{1}}^{2})({a_{2}}^{2}+{c_{2}}^{2})$ = ${(a_{1}a_{2}+c_{1}c_{2})}^{2}+ {(a_{1}c_{2}-c_{1}a_{2})}^{2}=m$.

We can continue in this way to show the claim if $m=p_{1}p_{2}...p_{k}$.

Let m be an integer satisfyng $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}=m$. $a,b,c,d \in \mathbb{Z}$, gcd(a,b)=1, gcd(c,d)=1.

$\frac{a^{2}d^{2}+b^{2}c^{2}}{b^{2}d^{2}}=m$

$a^{2}d^{2}+b^{2}c^{2}=mb^{2}d^{2}$

$b^{2}(md^{2}-c^{2})=a^{2}d^{2}$.

As our assumtion $b^{2} \not | a^{2}$ so $b^{2} | d^{2}$, $d^{2}=kb^{2}$. Then $\frac{1}{b^{2}}(a^{2}+\frac{c^{2}}{k})=m$.

$\frac{c^{2}}{k}=mb^{2}-a^{2}$ which is an integer. Thus we have $k=1$ as gcd(c,d)=1.

We have $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{b^{2}}=m$, $a^{2}+c^{2}=mb^{2}={p_{1}}^{\alpha_{1}}...{p_{k}^{\alpha_{k}}}$ [Unique prime factorization].

If $a^{2}+c^{2}=wp$, p is prime, then we can find integers x, y such that $x^{2}+y^{2}=p$.

Thus for every $p_{1}$, $p_{2}$,...,$p_{k}$, we can find $a_{1}, a_{2},...,a_{k}$ and $c_{1}, c_{2},...,c_{k}$ such that

${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$

${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$

$\vdots$

${a_{k}}^{2}+{c_{k}}^{2}=p_{k}$

Now if $m=p_{1}p{2}$, ${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$, ${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$.

$({a_{1}}^{2}+{c_{1}}^{2})({a_{2}}^{2}+{c_{2}}^{2})$ = ${(a_{1}a_{2}+c_{1}c_{2})}^{2}+ {(a_{1}c_{2}-c_{1}a_{2})}^{2}=m$.

If $m={p_{1}}^{2}$, then $m={({a_{1}}^{2}+{c_{1}}^{2})}{({a_{1}}^{2}+{c_{1}}^{2})}={({a_{1}}^{2}+{c_{1}}^{2})}^{2}+0$. We can continue in this way to show the claim.

This is an answer to the second question, "what is the simplest direct proof?"

Let m be an integer satisfyng $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}=m$. $a,b,c,d \in \mathbb{Z}$, gcd(a,b)=1, gcd(c,d)=1.

$\frac{a^{2}d^{2}+b^{2}c^{2}}{b^{2}d^{2}}=m$

$a^{2}d^{2}+b^{2}c^{2}=mb^{2}d^{2}$

$b^{2}(md^{2}-c^{2})=a^{2}d^{2}$.

As our assumtion $b^{2} \not | a^{2}$ so $b^{2} | d^{2}$, $d^{2}=kb^{2}$. Then $\frac{1}{b^{2}}(a^{2}+\frac{c^{2}}{k})=m$.

$\frac{c^{2}}{k}=mb^{2}-a^{2}$ which is an integer. Thus we have $k=1$ as gcd(c,d)=1.

We have $\frac{a^{2}}{b^{2}}+\frac{c^{2}}{b^{2}}=m$, $a^{2}+c^{2}=mb^{2}={p_{1}}^{\alpha_{1}}...{p_{k}^{\alpha_{k}}}$ [Unique prime factorization].

If $a^{2}+c^{2}=wp$, p is prime, then we can find integers x, y such that $x^{2}+y^{2}=p$.

Thus for every $p_{1}$, $p_{2}$,...,$p_{k}$, we can find $a_{1}, a_{2},...,a_{k}$ and $c_{1}, c_{2},...,c_{k}$ such that

${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$

${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$

$\vdots$

${a_{k}}^{2}+{c_{k}}^{2}=p_{k}$

Now if $m=p_{1}p{2}$, ${a_{1}}^{2}+{c_{1}}^{2}=p_{1}$, ${a_{2}}^{2}+{c_{2}}^{2}=p_{2}$.

$({a_{1}}^{2}+{c_{1}}^{2})({a_{2}}^{2}+{c_{2}}^{2})$ = ${(a_{1}a_{2}+c_{1}c_{2})}^{2}+ {(a_{1}c_{2}-c_{1}a_{2})}^{2}=m$.

If $m={p_{1}}^{2}$, then $m={({a_{1}}^{2}+{c_{1}}^{2})}{({a_{1}}^{2}+{c_{1}}^{2})}={({a_{1}}^{2}+{c_{1}}^{2})}^{2}+0$. We can continue in this way to show the claim.