Putnam 2024, Problem A-1

Determine all positive integers \( n \) for which there exist positive integers \( a, b, \) and \( c \) satisfying
\[ 2a^n + 3b^n = 4c^n. \]

Solution:
The answer is $n=1$. When $n=1$, $(a,b,c)=(1,2,2)$ is a solution to the given equation. We claim that there are no solutions when $n \geq 2$.

For $n=2$, suppose that we have a solution to $2a^2 + 3b^2 = 4c^2$ with $a,b,c \in \mathbb{N}$. By dividing each of $a,b,c$ by $\gcd(a,b,c)$, we obtain another solution; thus we can assume that $\gcd(a,b,c)=1$. Note that we have $a^2 + c^2 \equiv 0 \pmod{3}$, and that only $0$ and $1$ are perfect squares mod $3$; thus we must have $a^2 \equiv c^2 \equiv 0 \pmod{3}$. But then $a,c$ are both multiples of $3$; it follows from $b^2 = 12(c/3)^2 - 6(a/3)^2$ that $b$ is a multiple of $3$ as well, contradicting our assumption that $\gcd(a,b,c)=1$.

For $n \geq 3$, suppose that $2a^n + 3b^n = 4c^n$. As in the previous case, we can assume $\gcd(a,b,c)=1$. Since $3b^n = 4c^n - 2a^n$, $b$ must be even. We can then write $a^n + 2^{n-1} \cdot 3(b/2)^n = 2c^n$, and so $a$ must be even. Then $2^{n-1}(a/2)^n + 2^{n-2} \cdot 3(b/2)^n = c^n$, and $c$ must be even as well. This contradicts our assumption that $\gcd(a,b,c)=1$.