Find all triplets of non-zero natural numbers $(a, b, c)$ such that

$$
2021^{a}+4=3^{b} \times 5^{c}
$$
In a Diophantine equation, one always starts by looking at small values of the parameters. Here, it suffices to factorize $2021^{a}+4$ into prime factors, so we look at the small values of $a$. Here, when $a=1$, the equation becomes $2025=3^{b} \times 5^{c}$, so $(a, b, c)=(1,4,2)$ is a solution. Moreover, if $a \geqslant 2$, no one wants to factorize $2021^{a}+4$, so we stop here our exploration of small cases.
In a second step, one should either proceed with factorizations or look at the equation modulo a well-chosen number $n$ to reduce the set of possible values of $a, b$, and $c$. The numbers 3 and 5 are clearly prime, and our study of the case $a=1$ indicates that $2021=45^{2}-4=43 \times 47$. We note this relation in a corner of our scratch paper for later use.
Finally, using the Chinese Remainder Theorem, we reduce our equation modulo $n$ when $n$ is a power of a prime. For example:
$\triangleright n=1, n=2$ and $n=5$ do not provide any information;
$\triangleright n=3$ indicates that $a$ is odd;
$\triangleright n=4$ indicates that $b$ is even;
$\triangleright n=7$ does not provide any easy-to-exploit information;
$\triangleright n=8$ indicates, since $a$ is odd and $b$ is even, that $c$ is even.
We then set $\beta=b / 2$ and $\gamma=c / 2$. Moreover, two of the three terms in our equation are squares, which allows us to obtain the factorization

$$
43^{a} \times 47^{a}=2021^{a}=\left(3^{\beta} \times 5^{\gamma}\right)^{2}-2^{2}=\left(3^{\beta} \times 5^{\gamma}+2\right)\left(3^{\beta} \times 5^{\gamma}-2\right) .
$$

Since $3^{\beta} \times 5^{\gamma}+2$ and $3^{\beta} \times 5^{\gamma}-2$ are two odd integers whose difference is 4, they are coprime.
Only one of these two factors is divisible by 43, and only one is divisible by 47. Several cases are possible a priori:
$\triangleright$ if 47 divides $3^{\beta} \times 5^{\gamma}-2$, then $47^{a}$ divides $3^{\beta} \times 5^{\gamma}-2$, so $3^{\beta} \times 5^{\gamma}-2 \geqslant 47^{a} \geqslant 43^{a} \geqslant 3^{\beta} \times 5^{\gamma}+2$, which is impossible;
$\triangleright$ if 43 and 47 divide $3^{\beta} \times 5^{\gamma}+2$, then $3^{\beta} \times 5^{\gamma}+2=43^{a} \times 47^{a}$ and $3^{\beta} \times 5^{\gamma}-2=1$, which is impossible since $3^{\beta} \times 5^{\gamma}-2 \geqslant 3 \times 5-2=13$.
Thus, 47 divides $3^{\beta} \times 5^{\gamma}+2$ and 43 divides $3^{\beta} \times 5^{\gamma}-2$, so that $3^{\beta} \times 5^{\gamma}+2=47^{a}$ and $3^{\beta} \times 5^{\gamma}-2=43^{a}$. But then

$$
4=\left(3^{\beta} \times 5^{\gamma}+2\right)-\left(3^{\beta} \times 5^{\gamma}-2\right)=47^{a}-43^{a} \geqslant 47^{a}-43 \times 47^{a-1}=4 \times 47^{a-1}
$$

which shows that $a \leqslant 1$.
In conclusion, the only solution to the problem is $(a, b, c)=(1,4,2)$.
Comment from the graders The problem was difficult, but many students managed to score points thanks to intelligent reflections. Finding a particular solution is always a step forward, as it allows one to see what can or cannot be proven: here $(1,4,2)$ was a solution, so one could expect to prove that $b$ or $c$ was even, and it was futile to prove that $b$ or $c$ was odd without additional assumptions on $a$.

Let's recall that looking modulo a composite number is rarely intelligent: looking modulo 15, for example, serves no purpose, since the same observations can be made modulo 3 and 5, and since the calculations are simpler, it allows one to avoid making mistakes.
Here, the key to the problem was to question the parity of $b, c$ to obtain a square: obtaining squares in Diophantine equations is crucial, as it allows one to factorize and thus simplify the problem. However, once the equation is factorized, some forget to specify that the two terms are positive and even strictly greater than 1, which means one cannot immediately conclude that one of the factors is $47^{a}$ and the other is $43^{a}$. Moreover, it would be good to rigorously justify at the end that $47^{a}=43^{a}+4$ implies $a=1$. For example, one can invoke that $47^{2}>43^{2}+4$ and say that by immediate induction $47^{k}>43^{k}+4$ if $k \geqslant 2$.

## Senior Problems