What is the value of y?

Question

(1) x^2 - y^2 = 5
(2) x and y are each positive integers

Bunuel · Accepted Answer

The red part is not correct. From x^2 - y^2 = 5 we cannot say that x-y=1 and x+y=5 (or vise-versa) because for (1) we don't know whether x and y are integers. So, for example it's possible that x-y=10 and x+y=1/2. Even if we knew that x and y are integers, still from (x+y)(x-y)=5 it follows that x+y=5 and x-y=1 (or vise versa) OR x+y=-5 and x-y=-1 (or vise versa).

What is the value of y?

(1) x^2 - y^2 = 5. Infinitely many solutions exist for x and y. Not sufficient.

(2) x and y are each positive integers. Not sufficient.

(1)+(2) Since  x  and  y  are  positive integers  then  x+y=integer>0  and  x-y=integer  AND  x+y>x-y . Thus from  x^2 - y^2 =(x+y)(x-y)= 5  we'd have that  x+y=5  and  x-y=1 , from which it follows that  y=2 . Sufficient.

Answer: C.

WoundedTiger · Answer

Hi,

From 1, we get (x-y)(x+y)=5 i.e (x-y)=1,5 or (x+y)= 5,1...solving we get 2 values of y (one positive and one negative)

From 2 alone we get nothing

combining 1 & 2 we get X and Y as postive and we get one solution

shreerajp99 · Answer

Hi Bunuel,

It sounds stupid but can u tell me about this step:

X+y=5 and X-y=1. I always used to think that it will mean x+y=5 or x-y=5.

Bunuel · Answer

When we consider the two statements together we have that both  x  and  y  are  positive integers , thus  x+y=integer>0  and  x-y=integer  AND  x+y>x-y .

Next, we also have that  x^2 - y^2 =(x+y)(x-y)= 5 , so we have that the product of two multiple, x+y and x-y is equal to 5, a prime number. Since  x+y=integer>0  and  x-y=integer , then x+y must be 5 AND x-y must be 1: 5*1=5.

Hope it's clear.

bumpbot · Answer

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What is the value of y?

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