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

image
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.

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About mvtrinh

Retired high school math teacher.
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