The problem asks for all positive integers $n$ such that the equation $2a^n + 3b^n = 4c^n$ has a solution in positive integers $a, b, c$.
Let’s analyze the equation for small values of $n$.
Case $n=1$: The equation becomes $2a + 3b = 4c$. We need to find positive integers $a, b, c$ satisfying this. Consider $b=2$, then $2a + 6 = 4c$, or $a+3 = 2c$. If $a=1$, then $1+3 = 4 = 2c$, so $c=2$. Solution $(a, b, c) = (1, 2, 2)$. So, $n=1$ is a possible value.
Case $n=2$: The equation becomes $2a^2 + 3b^2 = 4c^2$. Consider modulo arithmetic. Modulo 3: $2a^2 \equiv 4c^2 \equiv c^2 \pmod{3}$. The squares modulo 3 are 0 and 1. If $a^2 \equiv 0 \pmod{3}$, then $a \equiv 0 \pmod{3}$. Then $0 \equiv c^2 \pmod{3}$, so $c \equiv 0 \pmod{3}$. If $a^2 \equiv 1 \pmod{3}$, then $2 \equiv c^2 \pmod{3}$, which is impossible. So, we must have $a = 3k$ and $c = 3m$ for some integers $k, m$. Substituting these into the equation: $2(3k)^2 + 3b^2 = 4(3m)^2$ $18k^2 + 3b^2 = 36m^2$ Dividing by 3: $6k^2 + b^2 = 12m^2$, so $b^2 = 12m^2 – 6k^2 = 6(2m^2 – k^2)$. This means $b^2$ is divisible by 6, so $b$ is divisible by 6 (since 6 is square-free times a square). Let $b = 6p$. Then $(6p)^2 = 6(2m^2 – k^2)$, so $36p^2 = 6(2m^2 – k^2)$, or $6p^2 = 2m^2 – k^2$. $k^2 = 2m^2 – 6p^2 = 2(m^2 – 3p^2)$. This means $k^2$ is even, so $k$ is even, $k = 2q$. $(2q)^2 = 2(m^2 – 3p^2)$, so $4q^2 = 2(m^2 – 3p^2)$, or $2q^2 = m^2 – 3p^2$. $m^2 = 2q^2 + 3p^2$.
We started with the equation $2a^2 + 3b^2 = 4c^2$. If there is a solution $(a, b, c)$, then $a, c$ are multiples of 3, and $b$ is a multiple of 6. If $(a, b, c)$ is a solution, then $(a/3, b/6, c/3)$ is also a solution to a related equation. Let $a = 3a_1, b = 6b_1, c = 3c_1$. $2(3a_1)^2 + 3(6b_1)^2 = 4(3c_1)^2$ $18a_1^2 + 108b_1^2 = 36c_1^2$ Dividing by 18: $a_1^2 + 6b_1^2 = 2c_1^2$. Modulo 3: $a_1^2 \equiv 2c_1^2 \pmod{3}$. If $c_1^2 \equiv 0$, then $a_1^2 \equiv 0$. If $c_1^2 \equiv 1$, then $a_1^2 \equiv 2$, impossible. So $a_1 = 3a_2, c_1 = 3c_2$. $(3a_2)^2 + 6b_1^2 = 2(3c_2)^2$ $9a_2^2 + 6b_1^2 = 18c_2^2$ Dividing by 3: $3a_2^2 + 2b_1^2 = 6c_2^2$. Modulo 3: $2b_1^2 \equiv 0 \pmod{3}$, so $b_1 \equiv 0 \pmod{3}$. Let $b_1 = 3b_2$. $3a_2^2 + 2(3b_2)^2 = 6c_2^2$ $3a_2^2 + 18b_2^2 = 6c_2^2$ Dividing by 3: $a_2^2 + 6b_2^2 = 2c_2^2$. This is the same form as the equation for $a_1, b_1, c_1$. By infinite descent, the only integer solution is $(0, 0, 0)$. Since we are looking for positive integer solutions, there are no such solutions for $n=2$.
Case $n \ge 3$: Consider modulo arithmetic. Modulo 2: $3b^n \equiv 0 \pmod{2}$, so $b^n \equiv 0 \pmod{2}$, so $b$ is even. Let $b = 2k$. $2a^n + 3(2k)^n = 4c^n$ $2a^n + 3 \cdot 2^n k^n = 4c^n$ $2a^n + 3 \cdot 2^n k^n = 2^2 c^n$
If $n=3$: $2a^3 + 24k^3 = 4c^3$, so $a^3 + 12k^3 = 2c^3$. $a^3$ is even, so $a$ is even. Let $a = 2m$. $(2m)^3 + 12k^3 = 2c^3$ $8m^3 + 12k^3 = 2c^3$ Dividing by 2: $4m^3 + 6k^3 = c^3$. $c^3$ is even, so $c$ is even. Let $c = 2p$. $4m^3 + 6k^3 = (2p)^3 = 8p^3$ Dividing by 2: $2m^3 + 3k^3 = 4p^3$. This is the original equation with $(m, k, p)$ instead of $(a, b, c)$. By infinite descent, the only integer solution is $(0, 0, 0)$. So, there are no positive integer solutions for $n=3$.
Consider the equation modulo some numbers. Modulo 3: $2a^n \equiv 4c^n \equiv c^n \pmod{3}$. If $a \not\equiv 0 \pmod{3}$ and $c \not\equiv 0 \pmod{3}$, then $a^n \equiv \pm 1 \pmod{3}$, $c^n \equiv \pm 1 \pmod{3}$. $2(\pm 1) \equiv \pm 1 \pmod{3}$. $2 \equiv 1$ or $2 \equiv -1 \equiv 2$, or $-2 \equiv 1$ or $-2 \equiv -1$. $2 \equiv 1$ (False), $2 \equiv 2$ (True), $-2 \equiv 1$ (True), $-2 \equiv -1$ (False). So, either $a^n \equiv c^n \pmod{3}$ or $a^n \equiv -c^n \pmod{3}$. If $n$ is even, $a^n \equiv 1, c^n \equiv 1$, then $2 \equiv 1$, false. If $n$ is even, $a^n \equiv 1, c^n \equiv 1$, $2(1) \equiv 1$, false.
If $n$ is odd, $a \equiv c \pmod{3}$ or $a \equiv -c \pmod{3}$. If $a \equiv 1, c \equiv 1$, $2 \equiv 1$, false. If $a \equiv 2, c \equiv 2$, $2(2) \equiv 2$, $4 \equiv 2$, $1 \equiv 2$, false. If $a \equiv 1, c \equiv 2$, $2(1) \equiv 2$, true. If $a \equiv 2, c \equiv 1$, $2(2) \equiv 1$, $4 \equiv 1$, $1 \equiv 1$, true.
Modulo $p$, a prime.
Consider the equation modulo a power of a prime. Let $v_p(x)$ be the exponent of the highest power of $p$ that divides $x$.
Consider $n \ge 2$. Assume there exists a solution $(a, b, c)$. Let $d = \gcd(a, b, c)$. Then $a = da’, b = db’, c = dc’$, where $\gcd(a’, b’, c’) = 1$. $2(da’)^n + 3(db’)^n = 4(dc’)^n$ $2d^n a’^n + 3d^n b’^n = 4d^n c’^n$ Dividing by $d^n$: $2a’^n + 3b’^n = 4c’^n$. So, we can assume $\gcd(a, b, c) = 1$.
If $n \ge 2$, and a solution exists, then $n=1$.
Consider the equation $2x + 3y = 4z$. Solutions are of the form $x = 2k – 3m, y = 4m – 2j, z = k$. If $x, y, z > 0$, then $2k > 3m$, $4m > 2j$, $k > 0$. We need to find positive integers $k, m, j$ satisfying these inequalities. Let $m=1$. $2k > 3$, $k \ge 2$. $4 > 2j$, $j = 1$. If $k=2, m=1, j=1$, then $x = 4 – 3 = 1, y = 4 – 2 = 2, z = 2$. $2(1) + 3(2) = 2 + 6 = 8$. $4(2) = 8$.
Consider the equation modulo small primes.
If $n$ is even, let $n = 2m$. $2a^{2m} + 3b^{2m} = 4c^{2m}$ $2(a^m)^2 + 3(b^m)^2 = 4(c^m)^2$. Let $A = a^m, B = b^m, C = c^m$. $2A^2 + 3B^2 = 4C^2$. We have shown that this equation has no solution in positive integers. So, if $n$ is even, there are no solutions. Thus, $n$ must be odd.
Assume $n \ge 3$ is odd, and there exists a solution. Modulo 4: $2a^n + 3b^n \equiv 0 \pmod{4}$. $2a^n – b^n \equiv 0 \pmod{4}$, so $2a^n \equiv b^n \pmod{4}$.
If $a$ is even, $a = 2k$, $2(2k)^n = 2^{n+1} k^n \equiv 0 \pmod{4}$ since $n+1 \ge 4$. Then $b^n \equiv 0 \pmod{4}$, so $b$ is even. If $a$ is odd, $a^n \equiv a \pmod{4}$. $2a \equiv b^n \pmod{4}$. If $a \equiv 1$, $2 \equiv b^n \pmod{4}$. If $n$ is odd, $b \equiv 2 \pmod{4}$. If $a \equiv 3$, $6 \equiv 2 \equiv b^n \pmod{4}$, so $b \equiv 2 \pmod{4}$.
Case 1: $a$ and $b$ are even. Let $a = 2a_1, b = 2b_1$. $2(2a_1)^n + 3(2b_1)^n = 4c^n$ $2 \cdot 2^n a_1^n + 3 \cdot 2^n b_1^n = 4c^n$ $2^{n+1} a_1^n + 3 \cdot 2^n b_1^n = 4c^n$ Divide by 4: $2^{n-1} a_1^n + 3 \cdot 2^{n-2} b_1^n = c^n$. If $n=3$, $4a_1^3 + 6b_1^3 = c^3$. Then $c$ is even, $c = 2c_1$. $4a_1^3 + 6b_1^3 = 8c_1^3$, $2a_1^3 + 3b_1^3 = 4c_1^3$. This leads to infinite descent.
Case 2: $a$ is odd, $b$ is even, $b = 2k$. $2a^n + 3(2k)^n = 4c^n$ $2a^n + 3 \cdot 2^n k^n = 4c^n$. Modulo 4: $2a^n \equiv 0 \pmod{4}$, impossible since $a$ is odd.
Consider the equation $2x^n + 3y^n = 4z^n$ in $\mathbb{Q}_p$. If there is a local solution for all primes $p$, then there might be a global solution.
Consider the equation modulo 5. $2a^n + 3b^n \equiv 4c^n \pmod{5}$. If $n=3$, $2a^3 + 3b^3 \equiv 4c^3 \pmod{5}$. Cubes modulo 5 are $0, 1, -1$. If $a^3, b^3, c^3 \in {1, -1}$, then $2(\pm 1) + 3(\pm 1) \equiv 4(\pm 1) \pmod{5}$. $\pm 2 \pm 3 \equiv \pm 4 \pmod{5}$.
Possibilities for $(\pm 2 \pm 3)$: $2+3=5 \equiv 0$, $2-3=-1$, $-2+3=1$, $-2-3=-5 \equiv 0$. Possibilities for $(\pm 4)$: $4, -4 \equiv 1$.
$0 \equiv 4$ (False), $0 \equiv 1$ (False) $-1 \equiv 4$ (False), $-1 \equiv 1$ (False) $1 \equiv 4$ (False), $1 \equiv 1$ (True) $0 \equiv 4$ (False), $0 \equiv 1$ (False)
So, we need $2a^3 \equiv -2 \pmod{5}$, $3b^3 \equiv 3 \pmod{5}$, $4c^3 \equiv 4 \pmod{5}$. $a^3 \equiv -1$, $b^3 \equiv 1$, $c^3 \equiv 1$. $a \equiv 4$, $b \equiv 1$, $c \equiv 1 \pmod{5}$. Example: $a=4, b=1, c=1$. $2(4^3) + 3(1^3) = 2(64) + 3 = 128 + 3 = 131$. $4(1^3) = 4$. $131 \not\equiv 4 \pmod{5}$.
Consider the equation $ax^n + by^n = cz^n$. If $n \ge 3$, by Fermat’s Last Theorem, $x^n + y^n = z^n$ has no positive integer solutions.
Consider the case when $n$ is odd. Assume there exists a solution $(a, b, c)$ for some odd $n \ge 3$. Let $(a, b, c)$ be a solution with the smallest value of $\max(a, b, c)$.
We found that if a solution exists, then $n=1$. We need to prove that if $n \ge 3$ is odd, there are no solutions.
Assume $n \ge 3$ is odd, and $(a, b, c)$ is a positive integer solution. Consider the equation modulo small numbers. Modulo 2: $b$ is even. Modulo 3: $2a^n \equiv c^n \pmod{3}$.
Let $n=3$. $2a^3 + 3b^3 = 4c^3$. We used infinite descent to show no solutions.
Assume $n \ge 3$ is odd. If a solution exists, let $(a, b, c)$ be a solution where $\gcd(a, b, c) = 1$. We know $b$ is even. If $a$ is even, then $4c^n$ is even, so $c$ is anything. If $a$ is odd, then $2a^n$ is $2 \pmod{4}$. $2 + 3b^n \equiv 0 \pmod{4}$, $2 – b^n \equiv 0 \pmod{4}$, $b^n \equiv 2 \pmod{4}$. If $n$ is odd, $b \equiv 2 \pmod{4}$.
If $a$ is even, $b$ is even. $2a^n + 3b^n = 4c^n$. $a = 2a_1, b = 2b_1$. $2^{n+1} a_1^n + 3 \cdot 2^n b_1^n = 4c^n$. Divide by 2: $2^n a_1^n + 3 \cdot 2^{n-1} b_1^n = 2c^n$. If $n \ge 2$, LHS is even, so RHS is even, $c$ is even. If $n=3$, $8a_1^3 + 12b_1^3 = 2c^3$, $4a_1^3 + 6b_1^3 = c^3$. $c$ is even, $c = 2c_1$. $4a_1^3 + 6b_1^3 = 8c_1^3$, $2a_1^3 + 3b_1^3 = 4c_1^3$.
Assume there is a solution for some odd $n \ge 3$. Consider the minimal solution in terms of $a+b+c$. If $a, b, c$ is a solution, then $a, b$ are even. $a = 2a_1, b = 2b_1, c = 2c_1$. $(a_1, b_1, c_1)$ is a solution to the same equation. $a_1 = a/2, b_1 = b/2, c_1 = c/2$. This contradicts minimality.
The only possibility is that the assumption of a solution for odd $n \ge 3$ is false. Therefore, the only positive integer $n$ for which a solution exists is $n=1$.
Final verification: If $n=1$, $2a + 3b = 4c$. Solution $(1, 2, 2)$. If $n=2$, no positive integer solutions by infinite descent. If $n \ge 3$ is odd, assume a solution exists. If $a, b, c$ is a solution, then $a, b$ are even, so $c$ is even. $(a/2, b/2, c/2)$ is a smaller positive integer solution, contradiction.
