The sum of two given fractions reduced to their lowest terms is an integer. Their denominators are positive. What is the maximum possible difference between their denominators?



Andrew B. solved this puzzle:

Let the fractions be \(\frac{a}{b}\) and \(\frac{c}{d}\), with \(b \leq d\) without loss of generality, and let their sum be \(n\). Then \(\frac{c}{d} = \frac{nb - a}{b}\), so \(\frac{c}{d}\) can be expressed as a fraction with \(b\) as the denominator. Therefore, \(b - d = 0\).

Hagen von Eitzen solved this puzzle:

The only possible difference is \(0\).

If \(\frac{a}{b} + \frac{c}{d} = n\) and \(\gcd(a, b) = \gcd(c, d) = 1\), then \(\frac{a}{b} = \frac{nd - c}{d}\). If \(e\) is a common divisor of \(nd - c\) and \(d\), then it is also a common divisor of \(c = nd - (nd -c)\) and \(d\). Hence \(\gcd(nd - c, d) = 1\), and by uniqueness of representation of a fraction in lowest terms, \(b = d\).

quasi solved this puzzle:

Suppose \(\frac{a}{b} + \frac{c}{d} = n\), where \(\frac{a}{b}\) and \(\frac{c}{d}\) are reduced fractions, \(b \gt 0\), \(d \gt 0\) and \(n\) is an integer.

Then \(\frac{c}{d} = n - \frac{a}{b}\). Therefore, \(\frac{c}{d} = \frac{bn - a}{b}\). Since \(a\) and \(b\) are relatively prime, \(bn - a\) and \(b\) are also relatively prime. Thus, \(\frac{bn - a}{b}\) is a reduced fraction, so \(\frac{c}{d} = \frac{bn - a}{b}\). It implies that \(c = bn - a\) and \(d = b\). Therefore, \(d - b = 0\).

Thomas Nordhaus solved this puzzle:

Let \(\frac{a}{b} + \frac{c}{d} = n\) with integers \(a\), \(b\), \(c\), \(d\) and \(n\), \(b \gt 0\), \(d \gt 0\), and \(\gcd(a, b) = gcd(c, d) = 1\). Cross-multiplying by \(bd\) results in \(ad + bc = nbd\). This can be written as \(bc = d(nb - a)\) or \(ad = b(nd - c)\). Since the right-hand sides are divisble by d and b, respectively, so are the left-hand sides. Therefore \(d \mid bc\) and \(b \mid ad\). Since \(\gcd(d, c) = 1\) and \(gcd(b,a) = 1\), it follows that \(d \mid b\) and \(b \mid d\). This is possible only if \(b = d\). Therefore, the maximum difference of the denominators is \(0\).

Corey Derochie solved this puzzle:

As per the problem statement, let \(\frac{a}{b} + \frac{c}{d} = n\), where \(a\), \(b\), \(c\), \(d\), and \(n\) are integers. In particular, \(b\) and \(d\) are positive integers. \(a\) and \(b\) are coprime to each other because \(\frac{a}{b}\) is in lowest terms. \(c\) and \(d\) are coprime to each other because \(\frac{c}{d}\) is in lowest terms.

From \(\frac{a}{b} + \frac{c}{d} = n\), we get \begin{equation} \label{eq1} b(nd - c) = ad. \end{equation} \(b\) does not equal \(0\) because \(\frac{a}{b}\) is defined. Suppose \((nd - c) = 0\). Then \(nd = c\). Thus, \(d\) divides \(c\). But \(c\) and \(d\) are coprime to each other, so \(d = 1\). Thus, \(c = n\). Necessarily, \(a = 0\) and \(b = 1\), to satisfy the constraints and \eqref{eq1}. Therefore, \(b\) divides \(d\).

If \((nd - c) \ne 0\), \(\gcd(a, b) = 1\), so \(a\) divides \((nd - c)\) by \eqref{eq1} and unique prime factorization. Thus, for some integer \(m\), \(ma = n*d - c\). Then \(bma = ad\), by \eqref{eq1}. Cancelling \(a\) on both sides we get \(bm = d\). Thus, \(b\) divides \(d\).

Similarly, we can show that \(d\) divides \(b\). Thus, \(b = d\) and \(b - d = 0\).

Siddhesh Chaubal solved this puzzle:

Let the two fractions in reduced form be \(\frac{n_1}{d_1}\) and \(\frac{n_2}{d_2}\). \(\newcommand{\lcm}{\operatorname{lcm}}\)

Let \(\lcm(d_1, d_2) = d\), and let \(d = x_1 d_1 = x_2 d_2\). The sum \(\frac{n_1}{d_1} + \frac{n_2}{d_2}\) is an integer, i.e. \(\frac{n_1 x_1 + n_2 x_2}{d}\) is an integer. Since \(x_1 \mid d\) and \(d \mid (n_1 x_1 + n_2 x_2)\), \(x_1 \mid (n_1 x_1 + n_2 x_2)\). This implies that \(x_1 \mid n_2 x_2\).

\(d = x_1 d_1 = x_2 d_2\) is the LCM of \(d_1\) and \(d_2\). If \(\gcd(x_1, x_2) = g \gt 1\), then \(d_1 \mid \frac{d}{g}\) and \(d_2 \mid \frac{d}{g}\), so \(\lcm(d_1, d_2) \lt d\). This is a contradiction, so \(\gcd(x_1, x_2) = 1\). Therefore, \(x_1 \mid n_2\).

Also, since \(x_1 \mid x_2 d_2\) and \(\gcd(x_1, x_2) = 1\), \(x_1 \mid d_2\). Since, \(\frac{n_2}{d_2}\) is in the lowest form, \(x_1 = 1\). By a similar argument, \(x_2 = 1\). Therefore, \(d = d_1 = d_2\). Thus, the answer is \(0\).