If $a$ and $b$ are coprime, prove that $a^2+b^2$ and $a^2b^2$ are coprime If $a$ and $b$ are coprime, prove that $a^2+b^2$ and $a^2b^2$ are coprime.
Answer:
We have to prove $\gcd(a^2+b^2,a^2b^2)=1$, $\gcd(a,b)=1$. Then $\gcd(a^2,b^2)=1$. Now we have $\gcd(a^2,a^2+b^2)=1$ and $\gcd(b^2,a^2+b^2)=1$ by $\gcd(a,b)=\gcd(p\pm q)$.
Then I am lost.

Suppose $(a,b)=1$. Bezout's Identity says there are $x,y\in\mathbb{Z}$ so that
$$
ax+by=1\tag1
$$
Then
$$
1=(ax+by)^3=\underbrace{\left(ax^3+3bx^2y\right)}_ua^2+\underbrace{\left(3axy^2+by^3\right)}_vb^2\tag2
$$
Rearranging $(2)$, we get
$$
u\left(a^2+b^2\right)+(v-u)b^2=1\tag3
$$
and
$$
(u-v)a^2+v\left(a^2+b^2\right)=1\tag4
$$
Multiplying $(3)$ and $(4)$ yields
$$
\left(\left(a^2+b^2\right)uv+\left(a^2u-b^2v\right)(u-v)\right)\left(a^2+b^2\right)-(u-v)^2a^2b^2=1\tag5
$$
Thus, $\left(a^2+b^2,a^2b^2\right)=1$.
