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1+2+⋯+10=12+22+⋯+521+2+⋯+13=12+22+⋯+62⋮\begin{aligned} 1 + 2 + \cdots + 10 &= 1^2 + 2^2 + \cdots + 5^2\\ 1 + 2 + \cdots + 13 &= 1^2 + 2^2 + \cdots + 6^2 \\ &\vdots \end{aligned}1+2+⋯+101+2+⋯+13=12+22+⋯+52=12+22+⋯+62⋮
Find the next pair of integers (M,N)(M,N)(M,N) larger than (13,6)(13,6)(13,6) such that 1+2+⋯+M=12+22+⋯+N2. 1 + 2 + \cdots +M= 1^2 + 2^2 + \cdots + N^2. 1+2+⋯+M=12+22+⋯+N2. Submit your answer as M+NM+NM+N.
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