1963 AHSME Problems/Problem 13

Problem

If $2^a+2^b=3^c+3^d$, the number of integers $a,b,c,d$ which can possibly be negative, is, at most:

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 0$

Solution

Assume $c,d \ge 0$, and WLOG, assume $a<0$ and $a \le b$. This also takes into accout when $b$ is negative. That means \[\frac{1}{2^{-a}} + 2^b = 3^c + 3^d\] Multiply both sides by $2^{-a}$ to get \[1 + 2^{-a+b} = 2^{-a} (3^c + 3^d)\] Note that both sides are integers. If $a \ne b$, then the right side is even while the left side is odd, so equality can not happen. If $a = b$, then $2^{-a} (3^c + 3^d) = 2$, and since $a<0$, $a = -1$ and $3^c + 3^d = 1$. No nonnegative value of $c$ and $d$ works, so equality can not happen. Thus, $a$ and $b$ can not be negative when $c,d \ge 0$.

Assume $a,b \ge 0$, and WLOG, assume $c < 0$ and $c \le d$. This also takes into account when $d$ is negative. That means \[2^a + 2^b = \frac{1}{3^{-c}} + 3^d\] Multiply both sides by $3^{-c}$ to get \[3^{-c} (2^a + 2^b) = 1 + 3^{d-c}\] That makes both sides integers. The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$, so equality can not happen. Thus, $c$ and $d$ can not be negative when $a,b \ge 0$.

Assume $a,c < 0$, and WLOG, let $a \le b$ and $c \le d$. This also takes into account when $b$ or $d$ is negative. That means \[\frac{1}{2^{-a}} + 2^b = \frac{1}{3^{-c}} + 3^d\] Multiply both sides by $2^{-a} \cdot 3^{-c}$ to get \[3^{-c} (1 + 2^{b-a}) = 2^{-a} (1 + 3^{d-c})\] That makes both sides integers. The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$, so equality can not happen. Thus, $a$ and $c$ can not be negative.

Putting all the information together, none of $a,b,c,d$ can be negative, so the answer is $\boxed{\textbf{(E)}}$.

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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