Question

The number of distinct pairs of integers (m, n) satisfying \(|1+mn| < |m+n| < 5\) is

Answer 1

Given the inequality:

\( |1 + mn| < |m + n| < 5 \)

Using the property \( |a| < |b| \iff a^2 < b^2 \) for real numbers \( a \) and \( b \):

\( (1 + mn)^2 < (m + n)^2 \)

Expanding both sides yields:

\( 1 + 2mn + m^2n^2 < m^2 + n^2 + 2mn \)

Subtracting \( 2mn \) from both sides:

\( 1 + m^2n^2 < m^2 + n^2 \)

Rearranging terms:

\( 1 - m^2 - n^2 + m^2n^2 < 0 \)

Factoring by grouping:

\( (1 - n^2) - m^2(1 - n^2) < 0 \)

Further factoring:

\( (1 - m^2)(1 - n^2) < 0 \)

For the product of two terms to be negative, one term must be positive and the other negative. This leads to two possibilities:

• \( 1 - m^2 > 0 \) and \( 1 - n^2 < 0 \), which implies \( |m| < 1 \) and \( |n| > 1 \).
• \( 1 - m^2 < 0 \) and \( 1 - n^2 > 0 \), which implies \( |m| > 1 \) and \( |n| < 1 \).

We now find integer solutions satisfying \( |m + n| < 5 \).

Case 1: \( m = 0 \) and \( |n| > 1 \)

The condition \( |m + n| < 5 \) becomes \( |n| < 5 \). Integers for \( n \) are \( \pm2, \pm3, \pm4 \), yielding 6 values.

Case 2: \( n = 0 \) and \( |m| > 1 \)

The condition \( |m + n| < 5 \) becomes \( |m| < 5 \). Integers for \( m \) are \( \pm2, \pm3, \pm4 \), yielding 6 values.

The total count of valid integer pairs \( (m, n) \) is 6 + 6 = 12