isn't it a trivial problem?when k=2022, take (a, b, c, d)=(p^674, q, q^674,p) now n=(p q)^675(just take p, q are primes) , and it's easy to show that when n=2023, there's no solution

China_BW wrote: isn't it a trivial problem?when k=2022, take (a, b, c, d)=(p^674, q, q^674,p) now n=(p q)^675(just take p, q are primes) , and it's easy to show that when n=2023, there's no solution You misread the problem. You solved when $(2022,n)$ is good while $(2023,n)$ is not.

thevictor wrote: China_BW wrote: isn't it a trivial problem?when k=2022, take (a, b, c, d)=(p^674, q, q^674,p) now n=(p q)^675(just take p, q are primes) , and it's easy to show that when n=2023, there's no solution You misread the problem. You solved when $(2022,n)$ is good while $(2023,n)$ is not. My fault,I'll gonna do it again

There exists a simpler construction but I'll post mine. Notice $(9x^4+3x)^3-(9x^3+1)^3=(9x^4)^3-1^3$. Let $x=3^{1011}$. Then, $(9x^4+3x)^3-(9x^3+1)^3=729^{2023}-1^{2023}$, so $$(9x^4+3x)^3+1^{2023}=(9x^3+1)^3+81^{2023}.$$Now, let $p$ be a sufficiently large prime. We get $$(p^{2023}(9x^4+3x))^3+(p^3)^{2023}=(p^{2023}(9x^3+1))^3+(81p^3)^{2023},$$so $(2023,3xp^{4052}(9x^3+1)(3x^3+1))$ is good. We claim that $(2022,3xp^{4052}(9x^3+1)(3x^3+1))$ is not good. We claim that $3xp^{4052}(9x^3+1)(3x^3+1)$ is not a square. Since $3xp^{4052}$ is a square and $\gcd(9x^3+1,3x^3+1)=2$, it suffices to show that $3x^3+1$ and $9x^3+1$ are not both two times a square. If $3^{3034}+1=2a^2$ and $3^{3035}+1=2b^2$, then $b^2-a^2=3^{3034}$, so since $a$ and $b$ are not multiples of $3$, we get $b-a=1$, contradiction. Now, let $k=3xp^{4052}(9x^3+1)(3x^3+1)$. Then, $\nu_p(k)=4052$. Since $k<p^{4052+\varepsilon}$, we can let $a=a_1p^{a_2}$, $b=b_1p^{b_2}$, $c=c_1p^{c_2}$, and $d=d_1p^{d_2}$ where $a_1^{2023},b_1^{2023},c_1^{2023},d_1^{2023}<p$. Since $675\nmid4052$, we cannot have $\nu_p(a^3)=\nu_p(b^{2022})=\nu_p(c^3)=\nu_p(d^{2022})$. Then, if $(2022,k)$ is good, we must have $\{a^3,b^{2022}\}=\{c^3,d^{2022}\}$, so either $a=c$ and $b=d$ or $a=d^{674}$ and $b=c^{674}$, which are both impossible since $k$ is not a square or a perfect $675$th power. Therefore, there exists infinitely many $n$ of the form $3xp^{4052}(9x^3+1)(3x^3+1)$ such that $(2022,n)$ is not good but $(2023,n)$ is good.