Try this beautiful problem based on Pairs of Positive Integer, useful for ISI B.Stat Entrance.
How many pairs of positive integers (m, n) are there satisfying \(m^3 – n^3 = 21\)?
Integer
Factorization
Odd number
Answer: (c) is an integral multiple of 6
TOMATO, Problem 156
Challenges and Thrills in Pre College Mathematics
Given that \(m^3– n^3 = 21\)
\(\Rightarrow (m – n)(m^2 + mn + n^2) = 3*7\)
Possible cases are , \(m – n = 3\), \(m^2 + mn + n^2 = 7\) and \(m – n = 1\), \(m^2 + mn + n^2 = 21\)
Can you now finish the problem ..........
Now according to the questions we are trying to find out the positive integers,so we will neglect the negetive cases..........
First case, \((3 + n)^2 + (3 + n)n + n^2 = 7\)
\(3 n^2 + 9n + 2 = 0\)
\(n = \frac {-9 \pm \sqrt {9^2 – 432}}{6}\)= not integer solution.
So this case is not possible.
Second case, \((n + 1)^2 + (n + 1)n + n^2 = 21\)
\(\Rightarrow 3n^2 + 3n – 20 = 0\)
\(n = \frac {-3 \pm \sqrt {9 + 240}}{6}\) = not integer solution.
Therefore option (b) is the correct
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