Belarus, TST, 2027

Rumination: This problem has an obvious line of attack. So, do not be surprised by what I am about to do. To know the why, refer to my book The Olympiad Companion, page number $\mathbf{3.47-3.48}. $

Solution: For any integer $x,$ we have the following:

Since $a^3, b^3 \in \{-1, 0, 1\} \pmod{7}$, and $20 \equiv -1 \pmod{7}$. For $20a^3 - b^3 \equiv 1 \pmod{7}$, we could have either one of the following two cases:

(A) $a^3 \equiv -1 \pmod{7}$ and $b^3 \equiv 0 \pmod{7}.$ Since $b$ is prime, therefore $b$ must be $7 \implies 20a^3 = 7^3 + 1 = 344,$ which does not have an integer solution. Therefore, this case is discarded.

(B) $a^3 \equiv 0 \pmod{7}$ and $b^3 \equiv -1 \pmod{7}.$ Since $a$ is prime, therefore $a$ must be $7.$ The equation then transforms into $20 \cdot 7^3 - 1 = b^3 \implies b = 19.$

So, the only primes $(a,b)$ that satisfy the given equation are $\mathbf{(7,19).}$

Another Solution: Since there is something like $b^3+1$, which can be factored, so factorisation might just work. Let us try! We have:

Therefore,

We first observe that $b\neq 2$, since $b=2$ would give

which is impossible. Hence, $b$ is odd.

Also,

If $b=3$, then

which is impossible. Therefore $b\neq3$, and hence

Now

Since $b$ is odd, $b^2-b+1$ is odd. Also, $b^2-b+1>5$ for $b\ge5$. Since the two factors $b+1$ and $b^2-b+1$ are coprime, it follows that

or

In the first case,

so $b=19$. Consequently,

giving $a=7$.

In the second case,

so $b=3$, which we have already ruled out.

Therefore, the only primes satisfying the given equation are

Indeed,

Remark: Notice that the second method would not have worked if the equation were something like $\mathbf{20a^3-b^3=2.}$ In that sense, the first solution was more general.