Number Theory

In refrence to the solution of Vishal S :-

Since a and b are natural numbers,

therefore a+b is greater than a-b.

So we should take only 3 pairs.

Let the two numbers be a,b

From given question

$a^2$ - $b^2$ =45

$\Rightarrow$ (a+b)(a-b)=45

By factorizing 45, we get

45

= $1 \times 45$

= $3 \times 15$

= $5 \times 9$

= $9 \times 5$

= $15 \times 3$

= $15 \times 1$

(a+b)(a-b)

= $1 \times 45$ **

= $3 \times 15$

= $5 \times 9$

= $9 \times 5$

= $15 \times 3$

= $15 \times 1$

From the above, we can form six pairs of equation

a+b=1 , a-b=45

a+b=3 , a-b=15

a+b=5 , a-b=9

a+b=45 , a-b=1

a+b=15 , a-b=3

a+b=9 , a-b=5

By solving the above 6 pairs of equation, we get

(a,b)=(23,22),(9,6),(7,2),(23,-22),(9,-6),(7,-2)

Since $-x^2$ = $x^2$ [I took x for an example]

Therefore we can take only magnitude in (23,-22),(9,-6) and (7,-2) i.e. (23,22),(9,6),(7,2)

The answer is (23,22),(9,6),(7,2)