Prove that for any positive integers $m>n$, there is infinitely many positive integers $a,b$ such that set of prime divisors of $a^m+b^n$ is equal to set of prime divisors of $a^{2019}+b^{1398}$.
Problem
Source: Iran MO 3rd round 2019 mid-terms - Number theory P2
Tags: number theory
kevinmathz
01.08.2019 21:23
Assume there are only a finite number that works. We can let $a^{2019}+b^{1398}=0$ mod p. Then, we observe that $a^{2019+(p-1)}+b^{1398}=0$ mod p due to FLT.
Now, all we have to do is prove that it works for all prime factors and only those prime factors.
Flash_Sloth
01.08.2019 22:00
It suffices to take an arbitrary integer $x$ and set \[ a = x^{n+1398}, b = x^{m+2019}\]Then \[ a^{2019}+b^{1398} = x^{2019(n+1398)} + x^{1398(m+2019)} = x^{1398*2019}(x^{2019n} +x^{1398m}) \]and \[ a^m +b^n = x^{m(n+1398)} +x^{n(m+2019)} = x^{mn}(x^{2019n} +x^{1398m}) \]Clearly they have the same set of prime divisors
guptaamitu1
27.06.2022 16:47
Flash_Sloth wrote:
It suffices to take an arbitrary integer $x$ and set
\[ a = x^{n+1398}, b = x^{m+2019}\]Then
\[ a^{2019}+b^{1398} = x^{2019(n+1398)} + x^{1398(m+2019)} = x^{1398*2019}(x^{2019n} +x^{1398m}) \]and
\[ a^m +b^n = x^{m(n+1398)} +x^{n(m+2019)} = x^{mn}(x^{2019n} +x^{1398m}) \]Clearly they have the same set of prime divisors
is motivated by trying to force \begin{align*} \frac{a^m}{b^{1398}} = \frac{b^n}{a^{2019}} = k \\ \implies \frac{a^m + b^n}{a^{2019} + b^{1398}} =k \end{align*}Now, $$ \frac{a^m}{b^{1398}} = \frac{b^n}{a^{2019}} \iff a^{m + 2019} = b^{1398 +n} $$So we take $$ a = x^{n+1398} ~,~ b = x^{m + 2019} $$and luckily the above construction just works.