## Product of Consecutive Integers

How many integer pairs $(m,n)$ satisfy the equation $m(m+1)=2^n$?
Source: NCTM Mathematics Teacher, September 2006

Solution
We construct products of positive consecutive integers $m(m+1)$ and see which ones if any can be expressed as a power of $2$. $1(2)=2=2^1$ $2(3)=6=2\cdot 3$ $3(4)=12=2^2\cdot 3$ $4(5)=20=2^2\cdot 5$ $5(6)=30=2\cdot 3\cdot 5$ $6(7)=42=2\cdot 3\cdot 7$ $7(8)=56=2^3\cdot 7$ $8(9)=72=2^3\cdot 3^2$ $9(10)=90=2\cdot 3^2\cdot 5$ $10(11)=2\cdot 5\cdot 11$ $\cdots$ $1(2)$ is the only such product and there is no other product because as the integers grow bigger and bigger, more and more prime numbers are introduced into the products making $m(m+1)=2^n$ impossible. $1(2)=1(1+1)=2^1$ yields the first solution $(m,n)=(1,1)$ and $-2(-1)=-2(-2+1)=2^1$ yields the second solution $(m,n)=(-2,1)$.

Answer: $2$

Alternative solution 1
We construct successive positive powers of $2$ and see which ones if any can be factored into a pair of factors that are different by $1$. $2^0=1=1(1)$ $2^1=2=1(2)$ $2^3=8=2(4)$ $2^4=16=2(8)=4(4)$ $\cdots$ $2^1=1(2)$ is the only such power.

Alternative solution 2 $m(m+1)=2^n$ $m^2+m-2^n=0$
Solving the quadratic equation by completing the square for $n=1$ $m^2+m-2^1=0$ $m^2+2m-m-2=0$ $m(m+2)-1(m+2)=0$ $(m+2)(m-1)=0$ $m=-2$ and $m=1$

Alternative solution 3 The figure shows the graphs of $y=x(x+1)$ and of $y=2^x$. They intersect at two points one of which has integer coordinates $(1,2)$ leading to $1(1+1)=2^1$. 