If x and y are positive real numbers, is x an integer? (1) x + y is an integer, and x^2 + … — GMAT

If $x$ and $y$ are positive real numbers, is $x$ an integer?

(1) $x + y$ is an integer, and $x^2 + xy$ is an integer.
(2) $x - y$ is an integer, and $x^2 - y^2$ is an integer.

A
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.Answer

Answer

Statements (1) and (2) together are not sufficient to determine whether

x

is an integer.

The correct option identifies that statements (1) and (2) together are not sufficient. Testing half-integers such as

x = 2.5

and

y = 1.5

shows that all given conditions (

x+y=4

x^2+xy=10

x-y=1

x^2-y^2=4

) are integers while

x

is not an integer. Conversely,

x = 3

and

y = 1

satisfies all conditions with an integer

x

. Thus, a definitive Yes/No answer cannot be determined.

Step-by-Step Solution

Analyze Statement (1) independently.

Statement (1) states that

x + y = k

(where

k

is an integer) and

x^2 + xy = x(x+y) = m

(where

m

is an integer). Thus,

x \cdot k = m \implies x = \frac{m}{k}

. Since

m

and

k

are integers,

x

is rational, but not necessarily an integer. For example, if

x = 1.5

and

y = 0.5

, then

x + y = 2

(integer) and

x^2 + xy = 1.5(2) = 3

(integer), but

x = 1.5

is not an integer. If

x = 2

and

y = 1

, then

x + y = 3

and

x^2 + xy = 6

, and

x = 2

is an integer. Thus, Statement (1) alone is INSUFFICIENT.

Variable

x

is only constrained to be a rational number

\frac{m}{k}

, which may or may not be an integer.

Analyze Statement (2) independently.

Statement (2) states that

x - y = d

(where

d

is an integer) and

x^2 - y^2 = (x-y)(x+y) = p

(where

p

is an integer). Thus

d(x+y) = p \implies x+y = \frac{p}{d}

. Then

2x = (x+y) + (x-y) = \frac{p}{d} + d = \frac{p + d^2}{d} \implies x = \frac{p + d^2}{2d}

. If

x = 2.5

and

y = 1.5

, then

x - y = 1

(integer) and

x^2 - y^2 = 6.25 - 2.25 = 4

(integer), but

x = 2.5

is not an integer. If

x = 2

and

y = 1

x - y = 1

and

x^2 - y^2 = 3

, and

x = 2

is an integer. Thus, Statement (2) alone is INSUFFICIENT.

Variable

x

is constrained to a rational form with denominator

2d

, which allows non-integer values.

Analyze Statements (1) and (2) combined.

Combining both statements gives

x + y = k

(integer) and

x - y = d

(integer). Adding these yields

2x = k + d

, so

2x

must be an integer. This means

x

can be an integer or a half-integer (e.g.,

2.5

). Testing

x = 2.5

and

y = 1.5

: both are positive real numbers;

x + y = 4

(integer);

x^2 + xy = 2.5(4) = 10

(integer);

x - y = 1

(integer); and

x^2 - y^2 = 6.25 - 2.25 = 4

(integer). Here

x

is NOT an integer. Testing

x = 3

and

y = 1

: both are positive real numbers;

x + y = 4

(integer);

x^2 + xy = 3(4) = 12

(integer);

x - y = 2

(integer); and

x^2 - y^2 = 9 - 1 = 8

(integer). Here

x

IS an integer. Because both 'Yes' and 'No' answers are possible under all conditions, Statements (1) and (2) together are INSUFFICIENT.

The system of equations allows

x

to be a half-integer (such as

2.5

), satisfying all integer conditions of both statements without requiring

x

itself to be an integer.

Key Concept

Evaluating implicit non-integer constraints and system degree in Data Sufficiency questions.

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