Problem 1 page 11 see

We see that either $p | (5^p-2^p)$ or $p | (5^q-2^q)$. Case 1: Let $p | (5^p-2^p)$ $$ 5^p\equiv 2^p \mod p \Rightarrow (5 \cdot2^{-1})^p\equiv 1 \mod p \Rightarrow 5\cdot2^{-1}\equiv 1\mod p$$But since we know that $2(p+1)\cdot 2^{-1}\equiv 1\mod p$ it follows that $$2(p+1)\equiv 5\mod p \Rightarrow 2p\equiv 3 \mod p \Rightarrow p=3$$ Now we could either have $q|5^3-2^3$ or $q|5^q-2^q$ $a$) Let $q|5^3-2^3$ Hence $q|3\times 3\times 13$ Since the solution pairs are symmetric w.r.t $p,q$ , the solutions are $(p,q)=(3,3);(3,13);(13,3)$ $b$) If $q|5^q-2^q$ we again just get $q=3$ giving us $(p,q)=(3,3)$ Case 2:Let $p | (5^q-2^q)$ and $q|(5^p-2^p)$. Let $5\cdot 2^{-1}=a$. $$5^q\equiv 2^q\mod p \Rightarrow a^q\equiv 1\mod p \Rightarrow ord_p(a)=q$$Similarly $ord_q(a)=p$ and hence, we have $q|p-1$ and $p|q-1$, which is impossible. Hence the only solutions are $(p,q)=(3,3);(3,13);(13,3)$.$\blacksquare$

Greenleaf5002 wrote: Case 2:Let $p | (5^q-2^q)$ and $q|(5^p-2^p)$. Let $5\cdot 2^{-1}=a$. $$5^q\equiv 2^q\mod p \Rightarrow a^q\equiv 1\mod p \Rightarrow ord_p(a)=q$$Similarly $ord_q(a)=p$ and hence, we have $q|p-1$ and $p|q-1$, which is impossible. Hence the only solutions are $(p,q)=(3,3);(3,13);(13,3)$.$\blacksquare$ Excuse me sir. I still confused about case 2. How can we know if $ord_p(a)=q$ and $ord_q(a)=p$ then, we have $q|p-1$ and $p|q-1$? Can anyone explain to me? Edit : Sorry, now I know. Case 2 : Let $p | (5^q-2^q)$ and $q|(5^p-2^p)$. Let $5\cdot 2^{-1}=a$ Assume $p \ne q$ If $a^{\color{red}{q}}\equiv 1\mod p$, $a^{\color{blue}{p}}\equiv 1\mod q$, and because p and q are prime, then p and q are the smallest number that satisfy the congruence of modulo. In mathematics, we can write $ord_p(a)=q$ and $ord_q(a)=p$. But by Fermat's Little Theorem : $a^{\color{red}{p-1}}\equiv 1\mod p$ and $a^{\color{blue}{q-1}}\equiv 1\mod q$. Then $\color{red}{q|p-1}$ and $\color{blue}{p|q-1}$ which is impossible (by write p = xq + 1 and q = yp + 1, we can conclude q may divide p-1 or p may divide q-1 but cannot happen simultaneously because $p \ne q$). Note : Order of $\color{red}{p-1}$ in FLT $a^{\color{red}{p-1}}\equiv 1\mod p$ is not the smallest number that satisfy the congruence. Example : By FLT : $2^{6}\equiv 1\mod 7$ But we know 3 is the smallest number that satisfy : $2^{3}=8\equiv 1\mod 7$

This is an obvious implication of LTE, isn't it?

IbrahimNadeem wrote: This is an obvious implication of LTE, isn't it? What is LTE?

It's a fundamental lemma for divisibility of $p^n - q^n$ for primes p,q. Namely, Lifting the Exponent