# Difference between revisions of "1979 AHSME Problems/Problem 22"

## Problem 22

Find the number of pairs $(m, n)$ of integers which satisfy the equation $m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1$.

$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }3\qquad \textbf{(D) }9\qquad \textbf{(E) }\infty$

## Solution

Solution by e_power_pi_times_i

Notice that $m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1 = \frac{(3n+1)^3}{27}$. Then $27(m^3 + 6m^2 + 5m) = (3n+1)^3$, and $(3n+1)^3 | 27$. However, $(3n+1)^3$ will never be divisible by $3$, nor $27$, so there are $\boxed{\textbf{(A)} 0 }$ integer pairs of $(m, n)$.

Solution 2: The equation is equivalent to $m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1$. Taking mod 3, we get $m(m+1)(m+2)=1 (\bmod 3)$. However, $m(m+1)(m+2)$ is always divisible by $3$ for any integer $m$. Thus, the answer is $\boxed{\textbf{(A)} 0}$ Solution by mickyboy789