2024 A1

A problem about raising integers to an arbitrary power.

Last updated 9/20/25

Problem Setup

Determine all positive integers \(n\) for which there exist positive integers \(a,b,\) and \(c\) satisfying \[\begin{equation*} 2a^n + 3 b^n = 4c^n \end{equation*}\]

Solution

For \(n=1\), brief trial and error will find that \(a=3\), \(b=2\), \(c=3\) is a solution: \[\begin{align*} 2(3)^1 + 3(2)^1 &= 4(3)^1 \\ 6+6 &= 12 \end{align*}\] For \(n=2\), we can assume that \(a,b,\) and \(c\) are relatively prime. Should there be a \(g = \text{gcd}(a,b,c)\), then we could divide out the common term a return the original problem: \[\begin{align*} 2(gx)^n + 3 (gy)^n &= 4 (gz)^n \qquad x,y,z \in \mathbb{N}\\ g^n(2x^n + 3y^n) &= g^n (4z^n) \\ 2x^n + 3y^n &= 4z^n \end{align*}\] We can then note that \(3b^2 = 4c^2 - 2a^2\) implies \[\begin{equation*} a^2+c^2 \equiv 0 \, \text{ mod(3)} \end{equation*}\] Furthermore, all perfect squares are either 0 (mod 3) or 1 (mod 3): \[\begin{align*} (3r)^2 &= 0 \text{ (mod 3)} \\ (3r+1)^2 &= (3r)^2 +2(3r) +1 = 3(r^3 +2r) +1 = 1 \text{ (mod 3)} \\ (3r +2)^2 &= (3r)^2 + 4(3r) + 4 = 3(r^2 +4r + 1) +1 = 1 \text{ (mod 3) } \end{align*}\] and so \(a^2 \equiv c^2 \equiv 0\) (mod 3), since neither \(a^2\) or \(c^2\) can be 1 (mod 3) and retain \(a^2+c^2 \equiv 0\) (mod 3). Therefore, \(a\) and \(c\) are multiples of 3. Then, \[\begin{align*} 3b^2 &= 4(3^2)\Big(\frac{c}{3} \Big)^2 - 2(3^2) \Big( \frac{a}{3} \Big)^2 \\ b^2 &= 12 \Big(\frac{c}{3} \Big)^2 - 6 \Big( \frac{a}{3} \Big)^2 \end{align*}\] which means that \(b\) is also a multiple of 3, which contradicts our assumption that \(a,b,c\) are relatively prime. There are no positive integers \(a, b,\) and \(c\) for which the given statement is true for \(n=2.\)

For the general case of \(n > 2\), we can note that \(b\) must be even, since \(3b^n = 4c^n - 2 a^n\), and \(3b^n\) is even only if \(b\) is even. We can rearrange the original equation to be of the form \[\begin{equation*} 2a^n = 4c^n-3b^n \end{equation*}\] which we can consider for the different parities of \(a\) and \(c\).

  1. If \(a\) and \(c\) are both even, the our original requirement of \(\text{gcd}(a,b,c) = 1\) is violated, since \(a,b,c\) all have a common factor of 2.

  2. If \(a\) and \(c\) are both odd, then for \(x,y,z \in \mathbb{N}\) we have \[\begin{align*} 2(2x+1)^n &= 4(2z+1)^n - 3(2y)^n \\[5pt] (2x+1)^n &= 2(2z+1)^n-2^{n-1}(3y^n) \end{align*}\] which cannot be true since the left side is odd and the right side is even.

  3. If \(a\) is odd and \(c\) is even, then \[\begin{align*} 2(2x+1)^n \neq 4(2z)^n - 3(2y)^n \end{align*}\] by the same argument as in the last case.

  4. Finally, if \(a\) is even and \(c\) is odd, we have \[\begin{align*} 2(2x)^n &= 4(2z+1)^n - 3(2y)^n \\[5pt] 2^{n-1}x^n &= (2z+1)^n - 2^{n-2}(3y^n) \end{align*}\] which, for \(n>2\), cannot be true since the right side is even and the left is odd.

Therefore, the problem statement is only true for \(n=1\).