AMC 10A 2024 (Problem 11)How many ordered pairs of integers (m,n)(m,n)(m,n) satisfy n2−49=m\sqrt{n^2-49}=mn2−49=m?(A) 1\text{(A)}\;1(A)1(B) 2\text{(B)}\;2(B)2(C) 3\text{(C)}\;3(C)3(D) 4\text{(D)}\;4(D)4(E) infinitely many\text{(E)}\;\text{infinitely many}(E)infinitely manyRelated TopicsCoreToolkit 10 — Difference of squaresCheck AnswerYour answer:ABCDECheckHints (6)Hint 1Square both sides.Hint 2Rearrange and use Toolkit 10 — Difference of squares.Hint 3n2−49=m\sqrt{n^2-49}=mn2−49=m⟹n2−49=m2\Longrightarrow n^2-49=m^2⟹n2−49=m2⟹n2−m2=49\Longrightarrow n^2-m^2=49⟹n2−m2=49⟹(n−m)(n+m)=49\Longrightarrow (n-m)(n+m)=49⟹(n−m)(n+m)=49Hint 4n−mn-mn−m and n+mn+mn+m are factors of 494949.Hint 5n2−49=m⟹m≥0\sqrt{n^2-49}=m\Longrightarrow m\ge0n2−49=m⟹m≥0⟹n−m≤n+m\Longrightarrow n-m\le n+m⟹n−m≤n+mHint 6By Hint 5, the cases aren=(n−m)+(n+m)2,m=(n+m)−(n−m)2n=\frac{(n-m)+(n+m)}{2},\qquad m=\frac{(n+m)-(n-m)}{2}n=2(n−m)+(n+m),m=2(n+m)−(n−m)n−mn+mnm14925247770−49−1−2524−7−7−70\begin{array}{c|c|c|c}n-m&n+m&n&m\\ \hline1&49&25&24\\7&7&7&0\\-49&-1&-25&24\\-7&-7&-7&0\end{array}n−m17−49−7n+m497−1−7n257−25−7m240240Final Answer(D) 444Related Problems (7)AMC 10A/12A Spring 2021 (Problem 10/9)AMC 10A 2012 (Problem 22)AMC 10B 2023 (Problem 14)BMO1 2016/2017 (Problem 3)AMC 10A/12A 2024 (Problem 15/9)AMC 10A 2023 (Problem 23)AMC 12B 2025 (Problem 3)