เกี่ยวกับ Number

[tatari/nightmare]

จงหาจำนวนเฉพาะ $p,q$ ทั้งหมดที่ทำให้ $pq\vert (5^p-2^p)(5^q-2^q)$ ช่วยคิดหน่อยครับ ขอบคุณมากครับ
นิป้า

[Erken]

Solution: The solutions are $(p, q)$ = (3, 3), (3, 13), or (13, 3). It is easy
to check that these are solutions.
Now we show that they are the only solutions. By symmetry, we may assume that $p \le q$.
Since $(5^p - 2^p)(5^q - 2^q )$ is odd, we have $q \ge p \ge 3.$
We observe that if a prime $k$ divides $5^k-2^k$ , then by Fermat’s little theorem,
we have $3 \equiv 5 - 2 \equiv 5^k - 2^k \pmod k$, or $k = 3.$
Assume that $p > 3$. By our observation, we have that $p$ divides $5^q - 2^q$ ,
or $5^q \equiv 2^q \pmod p$. By Fermat’s little theorem, we have $5^{p-1} \equiv 2^{p-1}
\pmod p$.
By Corollary 1.23,
$5^{\gcd(p-1,q)} \equiv 2^{\gcd(p-1,q)} \pmod p.$
Because $q \ge p$, $\gcd(p-1, q) = 1$. The last congruence relation now reads $5 \equiv 2 \pmod p$,
implying that $p = 3$, a contradiction. Hence $p = 3$.
If $q > 3$, by our observation, $q$ must divide $5^p - 2^p = 5^3 - 2^3 = 9 \cdot 13$, and so $q = 13$,
leading to the solution $(p, q)$ = (3, 13).

[nongtum]

แล้ว Corollary 1.23 มีใจความว่าอย่างไรหรือครับ

[Erken]

ขอโทษครับ
Corollary 1.23. Let m be a positive integer, and let a and b be integers relatively
prime to m. If x and y are integers such that
$a^x ≡ b^x (mod m)$ and $a^y ≡ b^y (mod m)$,
then
$a^{gcd(x,y)} ≡ b^{gcd(x,y)} (mod m)$