Question

The integer \( v \) is greater than 1. If \( v \) is the square of an integer, which of the following numbers must also be the square of an integer? Indicate all such numbers.

• A. \( 81v \)
• B. \( 25v + 10\sqrt{v} + 1 \)
• C. \( 4v^2 + 4\sqrt{v} + 1 \)
• D.
• E.

Hint:

When checking if an algebraic expression is a perfect square, try factoring it into the form \((ax + b)^2\). If it expands correctly, the expression is a square.

Answer 1

The Correct Option is A

Step 1: Assume \( v \) is a square.
Let \( v = k^2 \), where \( k \) is an integer and \( k>1 \).
Step 2: Check option (A).
\[ 81v = 81k^2 = (9k)^2, \] which is clearly a perfect square. Hence, (A) is correct.
Step 3: Check option (B).
\[ 25v + 10\sqrt{v} + 1 = 25k^2 + 10k + 1. \] Now observe that: \[ (5k + 1)^2 = 25k^2 + 10k + 1. \] Thus, this is also a perfect square. Hence, (B) is correct.
Step 4: Check option (C).
\[ 4v^2 + 4\sqrt{v} + 1 = 4k^4 + 4k + 1. \] This expression cannot be simplified into the square of an integer (for example, \((2k^2 + 1)^2 = 4k^4 + 4k^2 + 1\), which is different). Hence, (C) is not a perfect square.
Step 5: Conclusion.
The numbers that must be perfect squares are: \[ \boxed{\text{(A) and (B)}} \]