Let k be a non-negative integer such that the given fraction equals k². Multiplying both sides by the denominator 2n + 3, we obtain 7n - 12 = 2k²n + 3k², assuming 2n + 3 ≠ 0.

Collecting all terms involving n on one side gives n(7 - 2k²) = 3k² + 12. Since k is an integer, 2k² cannot equal 7, so 7 - 2k² ≠ 0. Thus, n can be expressed as a rational function of k².

To eliminate k² from the numerator, multiply n by 2: 2n = (6k² + 24)/(7 - 2k²). Rewriting the numerator as -3(7 - 2k²) + 45, we find that 2n = -3 + (45)/(7 - 2k²). For n to be an integer, 7 - 2k² must be a divisor of 45.

To eliminate k² from the numerator, multiply n by 2: 2n = (6k² + 24)/(7 - 2k²). Rewriting the numerator as -3(7 - 2k²) + 45, we find that 2n = -3 + (45)/(7 - 2k²). For n to be an integer, 7 - 2k² must be a divisor of 45.

When k = 1, n = 3. Substituting n = 3 back into the original expression: (7(3) - 12)/(2(3) + 3) = (9)/(9) = 1 = 1², which is a perfect square. When k = 2, n = -24. Substituting n = -24: (7(-24) - 12)/(2(-24) + 3) = (-180)/(-45) = 4 = 2², which is also a perfect square. Thus, the valid integer values are n = 3 and n = -24.

When k = 1, n = 3. Substituting n = 3 back into the original expression: (7(3) - 12)/(2(3) + 3) = (9)/(9) = 1 = 1², which is a perfect square. When k = 2, n = -24. Substituting n = -24: (7(-24) - 12)/(2(-24) + 3) = (-180)/(-45) = 4 = 2², which is also a perfect square. Thus, the valid integer values are n = 3 and n = -24.

Find all integer values of $n$ for which the expression $\frac{7n-12}{2n+3}$ results in a perfect square.

Answer: $n = 3$ and $n = -24$

Step-by-step solution

Step 1: Set Expression Equal to a Square and Clear Fraction

Let $k$ be a non-negative integer such that the given fraction equals $k^2$ . Multiplying both sides by the denominator $2n + 3$ , we obtain $7n - 12 = 2k^2n + 3k^2$ , assuming $2n + 3 \neq 0$ .

Step 2: Rearrange and Express n in Terms of k

Collecting all terms involving $n$ on one side gives $n(7 - 2k^2) = 3k^2 + 12$ . Since $k$ is an integer, $2k^2$ cannot equal $7$ , so $7 - 2k^2 \neq 0$ . Thus, $n$ can be expressed as a rational function of $k^2$ .

Step 3: Perform Polynomial Division

To eliminate $k^2$ from the numerator, multiply $n$ by $2$ : $2n = \frac{6k^2 + 24}{7 - 2k^2}$ . Rewriting the numerator as $-3(7 - 2k^2) + 45$ , we find that $2n = -3 + \frac{45}{7 - 2k^2}$ . For $n$ to be an integer, $7 - 2k^2$ must be a divisor of $45$ .

Step 4: Analyze Divisors of 45

We test integer values of $k \ge 0$ . Since $k^2 \ge 0$ , $7 - 2k^2 \le 7$ . Checking small non-negative integers $k$ : for $k = 0$ , $7 - 2(0) = 7$ (not a divisor of $45$ ); for $k = 1$ , $7 - 2(1) = 5$ (divides $45$ ); for $k = 2$ , $7 - 2(4) = -1$ (divides $45$ ); for $k = 3$ , $7 - 2(9) = -11$ (does not divide $45$ ); for $k = 4$ , $7 - 2(16) = -25$ (does not divide $45$ ); for $k = 5$ , $7 - 2(25) = -43$ ; for $k = 6$ , $7 - 2(36) = -65$ . As $k$ increases, $7 - 2k^2$ becomes strictly less than $-45$ , so no further divisors are possible.

Step 5: Solve for n and Verify

When $k = 1$ , $n = 3$ . Substituting $n = 3$ back into the original expression: $\frac{7(3) - 12}{2(3) + 3} = \frac{9}{9} = 1 = 1^2$ , which is a perfect square. When $k = 2$ , $n = -24$ . Substituting $n = -24$ : $\frac{7(-24) - 12}{2(-24) + 3} = \frac{-180}{-45} = 4 = 2^2$ , which is also a perfect square. Thus, the valid integer values are $n = 3$ and $n = -24$ .

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Find all integer values of nnn for which the expression 7n−122n+3\frac{7n-12}{2n+3}2n+37n−12​ results in a perfect square.