Data Sufficiency

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.
(2) x - y = 3

Mo2men wrote: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.
(2) x - y = 3

An odd integer can be expressed in the following form:
2k + 1.
Since y is odd, y = 2k + 1.

Statement 1:
In other words, p is 5 more than a multiple of 8:
p = 8a + 5.

Since p = x² + y² and p = 8a + 5, we get:
x² + y² = 8a + 5
x² + (2k+1)² = 8a + 5
x² + 4k² + 4k + 1 = 8a + 5
x² + 4k² + 4k = 8a + 4
x² = 8a + 4 - (4k²+4k)
x² = 4(2a+1) - 4k(k+1)
x² = 4[2a+1 - k(k+1)].

As shown above, 2a+1 = odd integer.
Since k(k+1) is the product of two consecutive integers, k(k+1) = even.
Thus, the resulting equation above implies the following:
x² = 4(odd - even)
x² = 4(odd)
x = 2(√odd)
x = 2(odd).

A multiple of 4 can be divided twice by 2.
Since x = 2(odd) CANNOT be divided twice by 2, x is NOT a multiple of 4.
SUFFICIENT.

Statement 2:
If x=4 and y=1, then x is a multiple of 4.
If x=5 and y=2, then x is NOT a multiple of 4.
INSUFFICIENT.

The correct answer is A.

GMATGuruNY wrote:
An odd integer can be expressed in the following form:
2k + 1.
Since y is odd, y = 2k + 1.

Statement 1:
In other words, p is 5 more than a multiple of 8:
p = 8a + 5.

Since p = x² + y² and p = 8a + 5, we get:
x² + y² = 8a + 5

Hi Mitch,

Thanks for your help

I have another solution for evaluating statement 1 based on plug in values carefully instead of using y=2k+1

x² + y² = 8a + 5
x² = 8a + 5 - y²

put y=1 & a = 1(odd number) x²= 4.....X= 2.............. Not divisible by 4

put y=1 & a = 2(even number) x²= 12.....X= 2 root 3...... Not divisible by 4

put y=3 & a = 3(odd number) x²= 20.....X= 2 root 10..... Not divisible by 4

put y=3 & a = 2(even number) x²= 12.....X= 2 root 3...... Not divisible by 4

put y=5 & a = 3(odd number) x²= 4.....X= 2.............. Not divisible by 4

We can conclude that a pattern will start to appear so Statement 1 is always 'NO'

sufficient.

Is my reasoning above valid??

Thanks

Mo2men wrote:

GMATGuruNY wrote:
An odd integer can be expressed in the following form:
2k + 1.
Since y is odd, y = 2k + 1.

Statement 1:
In other words, p is 5 more than a multiple of 8:
p = 8a + 5.

Since p = x² + y² and p = 8a + 5, we get:
x² + y² = 8a + 5

Hi Mitch,

Thanks for your help

I have another solution for evaluating statement 1 based on plug in values carefully instead of using y=2k+1

x² + y² = 8a + 5
x² = 8a + 5 - y²

put y=1 & a = 1(odd number) x²= 4.....X= 2.............. Not divisible by 4

put y=1 & a = 2(even number) x²= 12.....X= 2 root 3...... Not divisible by 4

put y=3 & a = 3(odd number) x²= 20.....X= 2 root 10..... Not divisible by 4

put y=3 & a = 2(even number) x²= 12.....X= 2 root 3...... Not divisible by 4

put y=5 & a = 3(odd number) x²= 4.....X= 2.............. Not divisible by 4

We can conclude that a pattern will start to appear so Statement 1 is always 'NO'

sufficient.

Is my reasoning above valid??

Thanks

Unfortunately, the three cases in red are not valid because x must be a positive integer.
The two cases in blue are valid, but they are insufficient to establish a pattern because x=2 in each case.
To establish a pattern, we need to find several different integer values for x that satisfy all of the constraints in the problem.

To test cases, make a list of options for perfect squares x² and y² and for 8a + 5:
x² and y²: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100...
8a + 5: 5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85, 93, 101, 109...

Now look for pairs in the first list that sum to a value in the second list:
1+4 = 5, implying that y=1 and x=2.
1+36 = 37, implying that y=1 and x=6.
9+36 = 45, implying that y=3 and x=6.
9+100 = 109, implying that y=3 and x=10.
The resulting values for x are all even non-multiples of 4.
Thus, it does not seem possible for x to be a multiple of 4.

Mo2men wrote: Since p = x² + y² and p = 8a + 5, we get:
x² + y² = 8a + 5

This is a nice idea, though. What you can do at this point is make y odd (= 2k + 1), then say

x² + (2k + 1)² = 8a + 5

x² + 4k² + 4k + 1 = 8a + 5

x² + 4k² + 4k = 8a + 4

x² = 8a + 4 - 4k² - 4k

Since x² = the sum of four multiples of 4, we know that x² is divisible by 4. Now let's suppose that x is a multiple of 4, that is that we can write x = 4z + 1, for some integer z. Plugging that in, we get

(4z + 1)² = 8a + 4 - 4k² - 4k

16z² + 8z + 1 = 8a + 4 - 4k² - 4k

or

Odd = Even

But that's impossible ... so x can't divide by 4, and we're done!