67th NMO Shortlisted Problems

Let us denote:

$$\begin{array}{rl}A& =n^2-4n+2\\ B& =n^2-3n+13\\ C& =n^2-6n+19\end{array}$$

We want $A$, $B$, $C$ to be simultaneously prime for some non-negative integer $n$.

Let us try small values of $n$:

For $n=0$: $$\begin{array}{rl}A& =0^2-4\cdot 0+2=2\\ B& =0^2-3\cdot 0+13=13\\ C& =0^2-6\cdot 0+19=19\end{array}$$ All are prime.

For $n=1$: $$\begin{array}{rl}A& =1-4+2=-1\\ B& =1-3+13=11\\ C& =1-6+19=14\end{array}$$ $A$ is negative, $C$ is not prime.

For $n=2$: $$\begin{array}{rl}A& =4-8+2=-2\\ B& =4-6+13=11\\ C& =4-12+19=11\end{array}$$ $A$ is negative.

For $n=3$: $$\begin{array}{rl}A& =9-12+2=-1\\ B& =9-9+13=13\\ C& =9-18+19=10\end{array}$$ $A$ is negative, $C$ is not prime.

For $n=4$: $$\begin{array}{rl}A& =16-16+2=2\\ B& =16-12+13=17\\ C& =16-24+19=11\end{array}$$ All are prime.

For $n=5$: $$\begin{array}{rl}A& =25-20+2=7\\ B& =25-15+13=23\\ C& =25-30+19=14\end{array}$$ $C$ is not prime.

For $n=6$: $$\begin{array}{rl}A& =36-24+2=14\\ B& =36-18+13=31\\ C& =36-36+19=19\end{array}$$ $A$ is not prime.

For $n=7$: $$\begin{array}{rl}A& =49-28+2=23\\ B& =49-21+13=41\\ C& =49-42+19=26\end{array}$$ $C$ is not prime.

For $n=8$: $$\begin{array}{rl}A& =64-32+2=34\\ B& =64-24+13=53\\ C& =64-48+19=35\end{array}$$ $A$ and $C$ are not prime.

For $n=9$: $$\begin{array}{rl}A& =81-36+2=47\\ B& =81-27+13=67\\ C& =81-54+19=46\end{array}$$ $C$ is not prime.

For $n=10$: $$\begin{array}{rl}A& =100-40+2=62\\ B& =100-30+13=83\\ C& =100-60+19=59\end{array}$$ $A$ is not prime.

For $n=11$: $$\begin{array}{rl}A& =121-44+2=79\\ B& =121-33+13=101\\ C& =121-66+19=74\end{array}$$ $C$ is not prime.

For $n=12$: $$\begin{array}{rl}A& =144-48+2=98\\ B& =144-36+13=121\\ C& =144-72+19=91\end{array}$$ None are prime.

Let us check if there is a pattern. Notice that for $n=0$ and $n=4$, all three are prime. Let's check $n=13$:

For $n=13$: $$\begin{array}{rl}A& =169-52+2=119\\ B& =169-39+13=143\\ C& =169-78+19=110\end{array}$$ None are prime.

It seems that only $n=0$ and $n=4$ work. Let's check if for $n\ge 5$ it is possible for all three to be prime.

For large $n$, $A$, $B$, $C$ are all roughly $n^2$ minus a linear term plus a constant. For $n$ even, $A$ is even, so unless $A=2$, it cannot be prime. For $n$ odd, $A$ is odd.

From above, for $n=0$, $A=2$ (prime), for $n=4$, $A=2$ (prime). For $n$ even and $n>4$, $A$ is even and greater than $2$, so not prime.

For $n$ odd, $A$ is odd, but from above, for $n=5$, $A=7$ (prime), but $C=14$ (not prime). For $n=7$, $A=23$ (prime), $C=26$ (not prime).

Therefore, the only non-negative integers $n$ for which all three expressions are simultaneously prime are $n=0$ and $n=4$.

Answer: $n=0$ and $n=4$.