Question

How many pairs of natural numbers are there such that the difference of their squares is 35?
1 , 2, 3, 4.

Answer 1

Answer:

There are only two pairs of natural numbers, difference of whose squares is 35.

Solution:

Let the two natural numbers be x, y

By the given condition,

x² - y² = 35 ( x > y, say)

or, x² = 35 + y² ..... (1)

In order to find the required pairs, we put y = 1, 2, 3, 4, ... where for a definite y, there must be an outcome x² which results a natural number x.

When y = 1, x² = 36 gives x = 6, a natural number

When y = 2, x² = 39 gives an irrational number

When y = 3, x² = 48 gives an irrational number

When y = 4, x² = 52 gives an irrational number

... ... ...

When y = 17, x² = 324 gives x = 18, a natural number

... ... ... and so on.

But the much higher values we input in calculations, the differences will be greater than 35 increasingly and we will have no such pair as our requirements.

∴ the only two pairs, difference of whose squares is 35 are (1, 6) and (17, 18).