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# Pairs of Positive Integer | ISI-B.stat | Objective Problem 178

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

Answer: (c) is an integral multiple of 6

TOMATO, Problem 156

Challenges and Thrills in Pre College Mathematics

## Try with Hints

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