x and y are positive integers, and x − y 0. If x + y has precisely

Question

x and y are positive integers, and  x  −  y  > 0. If  x  +  y  has precisely three positive divisors, and  x  −  y  has precisely two positive divisors, what is the smallest possible number of positive divisors that  x^2  −  y^2  could have?

A) 2

B) 3

C) 4

D) 6

E) 8

Abhishekgmat87 · Answer

Factor of x-y are 1 and x-y
x+y has 3 factors, 2 of which will be 1 and x+1. 
For smallest number of positive divisors, x-1 May be the 3rd factor of x+y.

So the product of x-y and x+y will have minimum 4 factors , 1, x-y, x+y, x^2-y^2

Take x=3,y=1,

x-y or 2 has 2 factors 1,2
x+y or 4 has 3 factors 1,2,4
x^2-y^2 or 8 will have 4 factors 1,2,4,8

Answer is 4 Option C

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SN_09 · Answer

Can someone please explain how to go about this question algebraically? In case, plugging in numbers takes time - want to know a better approach.

Also,  Abhishekgmat87  - can you please explain how you arrived at the below conclusion? I understood x+y has one factor as 1 and second factor as x+y. How is the third factor x-1 or x+1? I'm not following.

"x+y has 3 factors, 2 of which will be 1 and x+1. 
For smallest number of positive divisors, x-1 May be the 3rd factor of x+y."

jm2k5 · Answer

To do this algebraically we need to realise a few things:

x+y = (prime)^2 = only then it can have precisely 3 divisors.
x-y = prime only then it can have precisely 2 divisors.

So, let x+y = p1^2
and x-y = p2

where p1 and p2 are primes,

(x^2-y^2) = (x+y)(x-y) = p1^2 * p2

The number of divisors of the above expression is (2+1)(1+1) = 6.
But we will have minimum number of divisors when p1 = p2 = p

Then,

(x+y)(x-y) =p^3

This will have (3+1) = 4 divisors.

x and y are positive integers, and x − y > 0. If x + y has precisely

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