This was also P4 from the first Romanian TST from 2022. However, we had the problem with $k{}$ instead of 2021.

To get an idea for what's going on, one should really begin by solving the problem for a simple case like $2$! In $a^{2021} + b^{2023} = 2c^{2022} + 2d^{2024}$ it makes sense to try to obtain $a^{2021} = 2c^{2022}$ and $b^{2023} = 2d^{2024}$, so set $a = 2^x$, $b=2^y$, $c=2^z$, $d=2^t$ and solve $2^{2021x} = 2^{2022z+1}$ with $2^{2023y} = 2^{2024t+1}$. This is possible since $2021x = 2022z + 1$ and $2023y = 2024t + 1$ are solvable by Bezout's lemma. Now onto generalizing for any rational number $\frac{m}{n}$. Consider $a = m^{x_1}n^{x_2}$, $b = m^{y_1}n^{y_2}$, $c=m^{z_1}n^{z_2}$, $d=m^{t_1}n^{t_2}$ -- we wish to solve $m^{2021x_1}n^{2021x_2+1} + m^{2023y_1}n^{2023y_2+1} = m^{2022z_1+1}n^{2022z_2} + m^{2024t_1+1}n^{2024t_2}$. This is possible since all of $2021x_1 = 2022z_1+1$, $2021x_2+1 = 2022z_2$, $2023y_1 = 2024t_1 + 1$ and $2023y_2+1 = 2024t_2$ are solvable by Bezout's lemma. (One can simplify the argument by noticing that $x_2 = z_2 = y_2 = z_2 = 1$ work but whatever. It is also not hard to spot the exact values $x_1 = 2021$, $z_1 = 2020$, $y_1 = 2023$, $t_1 = 2022$.)

Can't you just do this? (or is it wrong) $a=x*t^{2023}$ $b=x*t^{2021}$ $c=x=d.$ For t containing the necessary prime divisors and x doing the rest of the job for the denominator.

This reminds me of my problem proposal which was published as Problem 1698 in Mathematics Magazine (proposed in Vol. 77, No. 3, June 2004 - solved in Vol 78, No. 3, June 2005) and was also posted here. ΜΜ1698. Let $n$ be an odd natural number. Show that every positive rational number can be written in the form \[ \dfrac{a_1^{n}+a_2^{n+1}+a_3^{n+2}}{b_1^{n}+b_2^{n+1}+b_3^{n+2}}, \]for some positive integers $a_i,b_i$ ($i=1,2,3$). Here is my write-up that accompanied my proposal to the Mathematics Magazine. Note that the above shortlist problem follows from the general case solved below: Solution. We shall prove more generally that if $n_i,m_i$ ($k\geq 1, 1\leq i \leq k$) are positive integers such that $(n_i,m_i)=1$ for $i=1,...,k$, then every positive rational number can be written in the form \[ \dfrac{a_1^{n_1}+a_2^{n_2}+...+a_k^{n_k}}{b_1^{m_1}+b_2^{m_2}+...+b_k^{m_k}}, \]for some positive integers $a_i,b_i$ ($i=1,...,k$). Indeed, first note that it suffices to consider the case $k=1$ (for $a/b=c/d$ implies that $a/b=c/d=(a+c)/(b+d)$). So consider two relatively prime natural numbers $n$ and $m$ and observe that there exist positive integers $\lambda, \mu, \lambda',\mu'$ such that \[ \mu n-\lambda m =1 \qquad \text{and} \qquad \lambda' m-\mu' n =1. \] Now put $\alpha=\lambda m=\mu n-1$, $\beta=\mu' n=\lambda' m-1$ and note that for the positive rational number $p/q$ we have \[ \dfrac{p}{q} = \dfrac{p^{\alpha +1} q^{\beta}}{p^{\alpha}q^{\beta +1}}=\dfrac{p^{\mu n}q^{\mu' n}}{p^{\lambda m}q^{\lambda' m}}=\dfrac{a^n}{b^m}, \] with $a=p^{\mu}q^{\mu'}$ and $b=p^{\lambda}q^{\lambda'}$. Finally, note that since $n$ is odd, we have $(n+2,n)=1=(n,n+1)=(n+1,n+2)$, and so our proposal follows from the more general result. Comments: (a). Although the above proposal follows from the more general case, we believe that it makes a harder problem. The general problem gives away the idea of our proof. (b) We can have several problems that follow from the general case. For example, if $p,q,r,s$ are four distinct primes, then every positive rational number can be written in the form \[ \dfrac{a^{p}+b^{q}}{c^{r}+d^{s}} \]for some positive integers $a,b,c,d$. In fact, the problem with $p=2,q=3,r=5$ and $s=7$ appeared in the 2nd round of the 51st Mathematical Olympiad (2000) in Poland. Other possible forms that follow from the above are: \[ \dfrac{a^{n}+b^{n+1}}{c^{n}+d^{n+1}}\,\text{(easier)} \qquad \text{and} \qquad \dfrac{a_1^{n}+a_2^{n+1}+...+a_{p+1}^{n+p}}{b_1^{m}+b_2^{m+1}+...+b_{p+1}^{n+p}}\,\text{(harder)}, \]where, in the first case, $n$ is any positive integer, and in the second $p$ is prime and $n$ is a positive integer not divisible by $p$. (The case $p=2$ is our proposal). (c) A slightly more difficult problem that does not follow from the above general case is that every positive rational number can be written in the form $(a^{3}+b^{3})/(c^{3}+d^{3})$ for some positive integers $a,b,c,d$. This is a problem from the 1999 IMO shortlist and recently appeared with a solution in Crux Mathematicorum.

This is a UK proposal by Dominic Yeo. The answer is yes. Let Set $a=x^{2019}, b=x^{2017}$ and $c=y^{2020}, d=y^{2018}$ for some integers $x,y$ and let $q=\frac{m}{n}$ in lowest terms. Then we could try to solve $$\frac{a^{2017}+b^{2019}}{c^{2018}+d^{2020}} = \frac{2x^{2017\cdot 2019}}{2y^{2018\cdot 2020}}=\frac{x^{2017\cdot 2019}}{y^{2018\cdot 2020}} = \frac{m}{n}.$$ Consider setting $x=m^{x_1}n^{x_2}$ and $y=m^{y_1}n^{y_2}$. Then by considering powers of $m$ and powers of $n$ separately, it would be sufficient to solve the pair of equations $$2017\cdot 2019 x_1 - 2018\cdot 2020 y_1 = 1 \quad\text{and}\quad 2017\cdot 2019 x_2 - 2018\cdot 2020 y_2 =-1. \quad \quad (\blacktriangle)$$ We know that these equations have solutions in positive integers if $2017\cdot 2019$ and $2018\cdot 2020$ are coprime. Amongst integers which differ by at most three, the only possible common prime factors are $2$ and $3$. Clearly $2$ does not divide the first, and we check that $3$ does not divide the second. So the integers are coprime, and the equations $(\blacktriangle)$ have solutions.

Let's set $a=k^{2021}m, b=k^{2023}m, c=k^{2020}m, d=k^{2022}m$ for any $k,m\in\mathbb{Z^+}$.Then we get $q=\dfrac{a^{2021}+b^{2023}}{c^{2022}+d^{2024}}=\dfrac{m^{2021}k^{2021^2}+m^{2023}k^{2023^2}}{m^{2022}k^{2021^2-1}+m^{2024}k^{2023^2-1}}=\dfrac{k}{m}*\dfrac{m^{2021}k^{2021^2-1}+m^{2023}k^{2023^2-1}}{m^{2021}k^{2021^2-1}+m^{2023}k^{2023^2-1}}=\dfrac{k}{m}$ Which means that $q$ can be represented as any positive rational number.

The answer is yes. Call a positive rational good if it can be represented in this way. Suppose $q = \tfrac{a^{2021}+b^{2023}}{c^{2022}+d^{2024}}$ is good. For any positive integer $r$, note that $\tfrac{q}{r}$ is also good, as it equals $\tfrac{(ra)^{2021}+(rb)^{2023}}{(rc)^{2022}+(rd)^{2024}}$. Thus, it suffices to prove that every positive integer is good. To do this, for an aribtrary positive integer $s$, let $a = s^w$, $b=s^x$, $c=s^y$, and $s=s^z$, where $w,x,y,z$ are positive integers. By CRT, we can choose $w,x,y,z$ such that $2021w = 2022y + 1$ and $2023x = 2024z+1$. Then, $s = \tfrac{(s^{w})^{2021}+(s^{x})^{2023}}{(s^{y})^{2022}+(s^{z})^{2024}}$ is good, so we are done. $\blacksquare$