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BMO1 2016/2017 (Problem 3)

Number TheoryMedium

Determine all pairs

(m,n)

of positive integers which satisfy the equation

n^2-6n=m^2+m-10

.

Core

Algebraic Identities

Your answer:

n^2-6n=(n-3)^2-9

m^2+m=\left(m+\frac12\right)^2-\frac14

By Hints 1 and 2,

n^2-6n=m^2+m-10

\Longrightarrow (n-3)^2-9=\left(m+\frac12\right)^2-\frac14-10

\Longrightarrow (2n-6)^2-36=(2m+1)^2-1-40

\Longrightarrow (2m+1)^2-(2n-6)^2=5

By Difference of squares,

(2m+1)^2-(2n-6)^2=5

\Longrightarrow (2m+1+2n-6)(2m+1-2n+6)=5

\Longrightarrow (2m+2n-5)(2m-2n+7)=5

2m+2n-5\text{ and }2m-2n+7\text{ are factors of }5

Let

A=2m+2n-5,\qquad B=2m-2n+7.

Then

m=\frac{A+B-2}{4},\qquad n=\frac{A-B+12}{4}.

The possible cases are

\begin{array}{c|c|c|c}A&B&m&n\\ \hline1&5&1&2\\[4pt]5&1&1&4\\[4pt]-1&-5&-2&4\\[4pt]-5&-1&-2&2\end{array}

Since

m

and

n

are positive integers,

(m,n)=(1,2),(1,4)

(m,n)=(1,2),(1,4)

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