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Difficulty: HardNumber Properties and Integer Constraints in Data Sufficiency

If xx is a positive real number, is xx an integer?

(1) x+1xx + \frac{1}{x} is an integer.

(2) x2+1xx^2 + \frac{1}{x} is an integer.

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

Answer

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
The option stating that both statements together are sufficient is correct. Evaluating Statement (1) alone allows irrational solutions like x=3+52x = \frac{3+\sqrt{5}}{2}. Evaluating Statement (2) alone allows irrational roots of x3−3x+1=0x^3 - 3x + 1 = 0. However, combining both statements reveals that x2−xx^2 - x is an integer. Subtracting the resulting linear system forces xx to be a rational number. For any positive rational x=pqx = \frac{p}{q} in simplest form, x+1xx + \frac{1}{x} being an integer requires p=1p = 1 and q=1q = 1, proving that x=1x = 1, which is an integer.

Step-by-Step Solution

1
Evaluate Statement (1) independently.
Statement (1) is NOT sufficient.
If x=1x = 1, then x+1x=2x + \frac{1}{x} = 2 (an integer), and xx is an integer. However, if x=3+52≈2.618x = \frac{3 + \sqrt{5}}{2} \approx 2.618 (a positive real non-integer), then x+1x=3+52+3−52=3x + \frac{1}{x} = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = 3 (an integer). Since xx can be an integer or a non-integer, Statement (1) alone does not uniquely answer the question.
2
Evaluate Statement (2) independently.
Statement (2) is NOT sufficient.
If x=1x = 1, then x2+1x=2x^2 + \frac{1}{x} = 2 (an integer), and xx is an integer. If x2+1x=3x^2 + \frac{1}{x} = 3, multiplying by xx yields x3−3x+1=0x^3 - 3x + 1 = 0. Evaluating f(x)=x3−3x+1f(x) = x^3 - 3x + 1 shows f(1)=−1<0f(1) = -1 < 0 and f(2)=3>0f(2) = 3 > 0, so by the Intermediate Value Theorem, there exists a real root x∈(1,2)x \in (1, 2), which is positive but not an integer. Hence, Statement (2) alone is insufficient.
3
Evaluate Statement (1) and Statement (2) together.
Both statements together are SUFFICIENT.
Let x+1x=ax + \frac{1}{x} = a and x2+1x=bx^2 + \frac{1}{x} = b, where aa and bb are integers. Subtracting Statement (1) from Statement (2) gives (x2+1x)−(x+1x)=b−a  ⟹  x2−x=k\left(x^2 + \frac{1}{x}\right) - \left(x + \frac{1}{x}\right) = b - a \implies x^2 - x = k, where k=b−ak = b - a is an integer. From Statement (1), x2−ax+1=0x^2 - ax + 1 = 0, and from the difference, x2−x−k=0x^2 - x - k = 0. Subtracting these two quadratic equations gives (1−a)x+(1+k)=0(1 - a)x + (1 + k) = 0. For any positive real xx, a=x+1x≥2a = x + \frac{1}{x} \ge 2, so 1−a≠01 - a \neq 0. Thus x=1+ka−1x = \frac{1 + k}{a - 1}. Because kk and aa are integers, xx must be a rational number. Let x=pqx = \frac{p}{q} in lowest terms, where pp and qq are positive integers with gcd⁡(p,q)=1\gcd(p, q) = 1. Then x+1x=pq+qp=p2+q2pq=a  ⟹  p2+q2=apqx + \frac{1}{x} = \frac{p}{q} + \frac{q}{p} = \frac{p^2 + q^2}{pq} = a \implies p^2 + q^2 = a p q. Since p2=q(ap−q)p^2 = q(ap - q), qq must divide p2p^2. But gcd⁡(p,q)=1\gcd(p, q) = 1, so q=1q = 1. Similarly, pp must divide q2q^2, so p=1p = 1. Thus x=1x = 1, which is an integer. Both statements together definitively answer YES.

Key Concept

Deduce integer constraints and rationality by combining non-linear algebraic expressions for real variables in Data Sufficiency.
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Topics

Number Properties and Integer Constraints in Data Sufficiency

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