72nd Czech and Slovak Mathematical Olympiad

First, let us note that at least one of the numbers $a+b$, $a+c$ and $b+c$ must be even. Indeed, two of the three numbers $a$, $b$, $c$ have the same parity, so their sum is even. If the product we investigate is a power of a prime $p$, then each of the four factors must be a power of $p$. As we already know, one of the first three factors is even, so $p=2$. Each of the four factors is therefore a power of two, which is greater than $1=2^0$, since the numbers $a,b,c$ are positive integers. Hence, each factor is an even number.

Further observe that the numbers $a+b$, $a+c$, $b+c$ are all even numbers, if and only if the numbers $a,b,c$ all have the same parity. Since $a+b+c+2036$ is even, $a,b$ and $c$ must be even numbers. Therefore, we can write $a=2a_1$, $b=2b_1$, and $c=2c_1$, where $a_1,b_1,c_1$ are positive integers. Then of course $$(a+b)(a+c)(b+c)(a+b+c+2036)=2^4(a_1+b_1)(a_1+c_1)(b_1+c_1)(a_1+b_1+c_1+1018).$$ The product of the last four parentheses must be a power of two. The numbers $a_1,b_1,c_1$ are therefore solutions to the original problem with the constant 2036 replaced by 1018. For the same reason as above $a_1,b_1$, and $c_1$ must be even. We denote by $a_2,b_2$ and $c_2$ respectively their halves (they are again positive integers) and we get $$(a+b)(a+c)(b+c)(a+b+c+2036)=2^8(a_2+b_2)(a_2+c_2)(b_2+c_2)(a_2+b_2+c_2+509).$$ And again, we have the same problem with the constant 509, so the numbers $a_2,b_2$, and $c_2$ necessarily have the same parity. However, since the number 509 is odd, $a_2,b_2$ and $c_2$ must be odd numbers. We see that the triple $a_2=b_2=c_2=1$ satisfies the problem (since $3+509=512$ is a power of two), so the corresponding triple $a=b=c=4$ is a solution to the original problem. We show that it is the only solution.

Suppose that at least one of the numbers $a_2,b_2,c_2$ is greater than one, let it be $c_2$ without loss of generality. Then, of course, the power of two equal to $a_2+c_2$ is greater than 2, so it is divisible by four. This means that dividing by four one of the numbers $a_2,c_2$ gives a remainder of 1 and the other a remainder of 3. So the third odd number $b_2$ has the same remainder when divided by four as one of the numbers $a_2,c_2$. The sum of $b_2$ with this number then has a remainder of 2, and since this sum is also a power of two, it must be the power of $2^1$. It follows that $b_2=1$ and $a_2=1$ (the equality of $c_2=1$ is ruled out by our assumption $c_2>1$). Thus the remainder 3 when divided by four necessarily gives $c_2$. But then the number $a_2+b_2+c_2+509$ gives remainder 2, so it is not a power of two, and that is a contradiction.

Conclusion. A single triple $(a,b,c)=(4,4,4)$ satisfies the problem statement.

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Alternative solution.

Let $a\ge b\ge c$ hold without loss of generality, so then $a+b\ge c+a\ge b+c\ge 2$. The product $(a+b)(b+c)(c+a)(a+b+c+2036)$ is a power of some prime number if and only if powers of that prime are all four factors of $a+b,b+c,c+a$, and $a+b+c+2036$. So let $a+b=p^k,c+a=p^l$, and $b+c=p^m$ for some prime $p$ and non-negative integers $k,l$ and $m$. For these, by the theorem of the introduction $k\ge l\ge m\ge 1$.

If $k>l$, and hence $k-1\ge l$ and $k-1\ge m$, with respect to $p\ge 2$, we would have $$a+b=p^k\ge p^{k-1}+p^{k-1}\ge p^l+p^m=(c+a)+(b+c)>a+b,$$ which is impossible. Necessarily, then, $k=l$, so $a+b=p^k=p^l=c+a$, where $b=c$. Then of course $p^m=b+c=2b$, and so $b=c=p^m/2$, so $2|p$, e.g. $p=2$, and therefore $b=c=2^{m-1}$ and $a+b=2^k$. Since $p=2$, it is also $a+b+c+2036=2^n$ for some integer $n$, which obviously satisfies the condition $2^n>2036$.

By the conclusion of the previous paragraph, the numbers $k,m,n$ satisfy the equality $$2^n=(a+b)+c+2036=2^k+2^{m-1}+2036.$$ Note that dividing by 16 number 2036 gives the remainder of 4, while $2^n$ certainly gives a remainder of 0 if $2^n>2036$. This implies $2^k+2^{m-1}\equiv 12(\mathrm{mod}16)$. The summands $2^k$ and $2^{m-1}$—as powers of two—are congruent to 1, 2, 4, 8 or 0 (mod 16). It is easy to see that the derived congruence holds only if one of the powers of $2^k,2^{m-1}$ yields a residue of 4 and the other 8. Then $k,m-1=2,3$. However, since $2^{m-1}=b<a+b=2^k$, it is necessarily $m-1<k$, and therefore $m-1=2$ and $k=3$. Hence $b=c=2^{m-1}=4$ and $a+b=2^k=8$, whence also $a=4$. This gives us the only possible triple $(a,b,c)=(4,4,4)$. This is indeed the solution to the problem—the product of the problem is then $8\cdot 8\cdot 8\cdot 2048$, while $8=2^3$ and $2048=2^{11}$.