If p, x, and y are positive integers, y is odd, and p = x^2

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: https://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).