BMO1 2016/2017 (Problem 3)Determine all pairs (m,n)(m,n)(m,n) of positive integers which satisfy the equation n2−6n=m2+m−10n^2-6n=m^2+m-10n2−6n=m2+m−10.Related TopicsCoreToolkit 10 — Difference of squaresMajorToolkit 6 — Square of a sumToolkit 7 — Square of a differenceHints (7)Hint 1n2−6n=(n−3)2−9n^2-6n=(n-3)^2-9n2−6n=(n−3)2−9Hint 2m2+m=(m+12)2−14m^2+m=\left(m+\frac12\right)^2-\frac14m2+m=(m+21)2−41Hint 3By Hints 1 and 2,n2−6n=m2+m−10n^2-6n=m^2+m-10n2−6n=m2+m−10⟹(n−3)2−9=(m+12)2−14−10\Longrightarrow (n-3)^2-9=\left(m+\frac12\right)^2-\frac14-10⟹(n−3)2−9=(m+21)2−41−10⟹(2n−6)2−36=(2m+1)2−1−40\Longrightarrow (2n-6)^2-36=(2m+1)^2-1-40⟹(2n−6)2−36=(2m+1)2−1−40⟹(2m+1)2−(2n−6)2=5\Longrightarrow (2m+1)^2-(2n-6)^2=5⟹(2m+1)2−(2n−6)2=5Hint 4By Toolkit 10 — Difference of squares,(2m+1)2−(2n−6)2=5(2m+1)^2-(2n-6)^2=5(2m+1)2−(2n−6)2=5⟹(2m+1+2n−6)(2m+1−2n+6)=5\Longrightarrow (2m+1+2n-6)(2m+1-2n+6)=5⟹(2m+1+2n−6)(2m+1−2n+6)=5⟹(2m+2n−5)(2m−2n+7)=5\Longrightarrow (2m+2n-5)(2m-2n+7)=5⟹(2m+2n−5)(2m−2n+7)=5Hint 52m+2n−5 and 2m−2n+7 are factors of 52m+2n-5\text{ and }2m-2n+7\text{ are factors of }52m+2n−5 and 2m−2n+7 are factors of 5Hint 6LetA=2m+2n−5,B=2m−2n+7.A=2m+2n-5,\qquad B=2m-2n+7.A=2m+2n−5,B=2m−2n+7. Thenm=A+B−24,n=A−B+124.m=\frac{A+B-2}{4},\qquad n=\frac{A-B+12}{4}.m=4A+B−2,n=4A−B+12. The possible cases areABmn15125114−1−5−24−5−1−22\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}A15−1−5B51−5−1m11−2−2n2442Hint 7Since mmm and nnn are positive integers,(m,n)=(1,2),(1,4)(m,n)=(1,2),(1,4)(m,n)=(1,2),(1,4)Final Answer(m,n)=(1,2),(1,4)(m,n)=(1,2),(1,4)(m,n)=(1,2),(1,4)Related Problems (10)AMC 10A 2012 (Problem 22)AMC 10B 2023 (Problem 14)AMC 10A 2024 (Problem 11)AMC 10/12A Spring 2021 (Problem 10)AMC 10A/12A 2024 (Problem 15/9)AMC 10A 2023 (Problem 23)AMC 10A 2020 (Problem 14)AMC 10A 2020 (Problem 5)View all related problems →