Try this beautiful problem based on Pairs of Positive Integer, useful for ISI B.Stat Entrance.

## Pairs of Positive Integer | ISI B.Stat Entrance | Problem 178

How many pairs of positive integers (m, n) are there satisfying \(m^3 – n^3 = 21\)?

- (a) Exactly one
- (b) None
- (c) Exactly three
- (d) Infinitely many

**Key Concepts**

Integer

Factorization

Odd number

## Check the Answer

But try the problem first…

Answer: (c) is an integral multiple of 6

TOMATO, Problem 156

Challenges and Thrills in Pre College Mathematics

## Try with Hints

First hint

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 ……….

Final Step

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

## Other useful links

- https://www.cheenta.com/pattern-problem-amc-10a-2003-problem-23/
- https://www.youtube.com/watch?v=PIBuksVSNhE

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