If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5 --> p=8q+5=x^2+y^2 --> as given that y=odd=2k+1 --> 8q+5=x^2+(2k+1)^2 --> x^2=8q+4-4k^2-4k=4(2q+1-k^2-k).

So, x^2=4(2q+1-k^2-k). Now, if k=odd then 2q+1-k^2-k=even+odd-odd-odd=odd and if k=even then 2q+1-k^2-k=even+odd-even-even=odd, so in any case 2q+1-k^2-k=odd --> x^2=4*odd --> in order x to be multiple of 4 x^2 must be multiple of 16 but as we see it's not, so x is not multiple of 4. Sufficient.

(2) x – y = 3 --> x-odd=3 --> x=even but not sufficient to say whether it's multiple of 4.

Answer: A.
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Such questions can be easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say if we need to divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For more on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibility-and-remainders-if-you.html

Coming to your question,

First thing that comes to mind is if y is odd, y^2 is also odd.
If y = 2k+1,
y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1
Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. So when y^2 is divided by 8, it will leave a 1.


Stmnt 1: When p is divided by 8, the remainder is 5.
When y^2 is divided by 8, remainder is 1. To get a remainder of 5, when x^2 is divided by 8, we should get a remainder of 4.
x^2 = 8a + 4 (i.e. we can make 'a' groups of 8 and 4 will be leftover)
x^2 = 4(2a+1) This implies x = 2*\sqrt{Odd Number}because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that x is not divisible by 4. Sufficient.

Stmnt 2: x - y = 3
Since y is odd, we can say that x will be even (Even - Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient.

Answer (A).
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Its not something one shall already know before attacking a question, you may realize properties like this when u start solving a question.
Even I didn't know about this property of 8.

I approached the question in following way:

Stmt 1: P/8=(x^2+y^2)/8; using remainder theorem;
rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8]
if x is divisible by 4, then x^2= 4k*4k= 16K=8*2K is also divisible by 8.
now to anaylze rem[y^2/8]; start putting suitable values of y; i.e all odd values starting from 1.
for y=1; rem(1/8)=1
for y=3; rem(9/8)=1
for y=5;rem(25/8)=1

so you observe this pattern here.
coming back to ques now, as rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8]=
rem[x^2/8] + 1 =5; this means rem[x^2/8] is not 0; which implies x is not divisible my 8;
Sufficient

Stmt2:
y being odd can be accept both 3 and 5 as values and we get different results; thus
Insufficient

Thus OA is A
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Am i missing something, why cant we take stmt 2 as follows:
squaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4

If p, x, and y are positive integers, y is odd, and p = x2 + y2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5. (2) x – y = 3.

In my opinion the answer should be D, please provide explanations with your responses. Thanks!

good ques thanks!!