Question

how many pairs of natural numbers are there if the difference of their squares is 60

Answer 1

a^2 - b^2 = 60

(a+b)(a-b) = (2^2) * 3 * 5

Guess you'll have to take cases here. And compare LHS and RHS ofcorse.

Answer 2

Note that 60 has 12 factors!

Answer 3

So you'll have to take 6 cases.

60 = 1*60 = 2*30 = 3*20 = 4*15 = 5*12 = 6*10
Does it help ?

Answer 4

no i want clearly because the answer is given 2

Answer 5

the only possible cases are \((a-b)=2,(a+b)=30\text{ and } (a-b)=6,(a+b)=10\)

Answer 6

See, you need to compare
First thing to keep in mind is a+b will be greater than a-b given that both are natural numbers

If a+b is odd, then a-b will also be odd
If a+b is even, then a-b will also be even

Keep those in mind and compare

Answer 7

as sire @sirm3d says.

Answer 8

if 3m = 3x and 3n = 27x then m equals to what?

Answer 9

m = log(e^x)

Answer 10

where base of log is e