How many pairs of natural numbers are there so that difference of their squares is 60?

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Solution

The correct option is C 2
$L e t t h e n u m b e r s b e x a n d y t h e n x^{2} - y^{2} = 60 \Rightarrow (1)$
$o r (x - y) (x + y) = 60 \Rightarrow (2)$

$F a c t o r s o f 60 :$
$5 \times 12$
$2 \times 30$
$4 \times 15$
$1 \times 60$
$3 \times 20$
$6 \times 10$

$W e g o f o r e v e n p a i r s (2, 30) a n d (6, 10)$

$s i n c e e v e n - o d d p a i r w o n t s a t i s f y c o n d i t i o n (2)$

$c h e c k i n g e v e n p a i r s i n c o n d i t i o n (1)$

$x - y = 2, x + y = 30 \to x = 16, y = 14$

$& x - y = 6, x + y = 10 \to x = 8, y = 2$

$H e n c e, o n l y 2 s e t s o f N a t u r a l n u m b e r s$

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