Question

If 2 positive integers M and N satisfy M² - N² = 2011 find M and N

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Answer 1

Answer:

Note that MN>1 since M≠N. Let p be any prime divisor of MN, and let pa and pb denote the highest power of p dividing M and N respectively. Thus the highest power of p dividing lcm(M,N)=max{a,b}, whereas p2⋅min{a,b} divides each of the terms M2, N2, MN.

If a=b, then the highest power of p dividing the LHS is a whereas p2a divides the RHS. This is only possible when a=b=0, which is impossible since p must divide at least one of M, N.

Hence a≠b. Equating highest powers of p dividing the two sides now gives max{a,b}=2⋅min{a,b}. But then a+b=max{a,b}+min{a,b}=3⋅min{a,b}, which means that the highest power of p dividing MN (which is a+b) is a multiple of 3. Since this is true of each prime divisor of MN, we conclude that MN is a perfect cube. ■

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Step-by-step explanation: