Let $A_p=a^{2^p}+b^{2^p}$. Prove that $(A_p,A_q)=1\forall p\ne q$

Question

Let $a,b\in N^*$ such that $2\mid ab$ and $(a,b)=1$. Consider $A_p=a^{2^p}+b^{2^p}$ show that $$(A_p,A_q)=1\forall p\ne q$$

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Let $(A_p,A_q)=d(d\in \mathbb{Z})$ then i will prove $d=1$

From $2\mid ab$ and $(a,b)=1$ we can assume $$a=2k,b=2k+1 (k\in \mathbb{N^*})$$

We have: $$d\mid (A_p-A_l)=\left(2k\right)^{2^p}+\left(2k+1\right)^{2^p}-\left(2k\right)^{2^l}-\left(2k+1\right)^{2^l}$$

note that: $\left(2k\right)^{2^p}-\left(2k\right)^{2^l}=2^{2^p}\cdot k^{2^p}-2^{2^l}\cdot k^{2^l}=1$. By lemma $1=(a,b)\Leftrightarrow \exists ax+by=1$ and $(F_k,F_l)=1\forall k\ne l(F_k=2^{2^k})$

Then $d\mid 1+\left(2k+1\right)^{2^p}-\left(2k+1\right)^{2^l}$

Im have no idea to solve $\left(2k+1\right)^{2^p}-\left(2k+1\right)^{2^l}$. Help me

Answer 1

Let $p<q$, and suppose a prime $r$ divides both $A_p$ and $A_q$ (it is then odd). The latter means $$a^{2^p}\equiv-b^{2^p}\pmod{r},\qquad a^{2^q}\equiv-b^{2^q}\pmod{r},$$ but, raising the first of these congruences into $2^{q-p}$-th power, we get $a^{2^q}\equiv b^{2^q}\pmod{r}$. With the second congruence, this gives $2a^{2^q}\equiv2b^{2^q}\equiv0\pmod{r}$, i.e. $r$ must divide $a$ and $b$. Contradiction.