Find all primes $p$ and $q$ such that $p^2+1 | 2003^q+1$ and $q^2+1 |2003^p+1$.

Question

Find all primes $p$ and $q$ such that $p^2+1 | 2003^q+1$ and $q^2+1 |2003^p+1$.

If $$p=2$$ then $$p^2+1=5$$ we have $$2003^q+1\equiv 0 [5]$$ $$2003^{2q}-1\equiv 0 [5]$$ let $o_{5}(2003)$ be the order of $2003$ modulo $5$

we have $$o_{5}(2003) |2q$$ $$2003\equiv 3[5]$$ and $$3^4\equiv 1[5]$$ $$o_{5}(2003)=4$$ so $$2 |q$$ Hence $$q=2$$ So $$(p,q)=(2,2)$$ Let's assume that $$(p,q)\neq(2,2)$$ I think we will use p-adic valuation. But I don't know how . Moreover , I m sure that there are no other solutions . But I need to prove it .

Answer 1

Here is a solution that i found in art of problem solving