December 28, 2024
Under division with remainder, \(a^2 + b^2\) divided by \(a+b\) yields a quotient \(q\) and a remainder \(r\). It is known that \(q^2 + r = 2005\). Find the minimum value of \(ab\), where \(a,b,q,r\) are all positive integers.
带余除法之下，\(a^2 + b^2\) 除以 \(a+b\) 得商和余数分别为\(q\)和\(r\)， 已知 \(q^2 + r = 2005\)， 求\(ab\)的最小值。（\(a,b,q,r\)均为正整数）
By setting up equations below: \[ a^2 + b^2 = q(a+b) + r \] \[ q^2 + r = 2005 \] and canceling \(r\), we get \[ (a-\frac{q}{2})^2 + (b-\frac{q}{2})^2 + \frac{q^2}{2} = 2005 \] Applying Cauchy–Schwarz inequality: \[ ((a-\frac{q}{2}) + (b-\frac{q}{2}) + \frac{q}{2} + \frac{q}{2})^2 \le (1+1+1+1)\cdot((a-\frac{q}{2})^2 + (b-\frac{q}{2})^2 + (\frac{q}{2})^2 + (\frac{q}{2})^2)\] we get \[ a + b \le 2 \sqrt{2005}. \]
Note that we also have \(0 \le r = 2005 - q^2 < a+b\), hence \[ 0 \le 2005 - q^2 \le 2 \sqrt{2005} \] which yields \(q=44\) as the only positive integer solution for \(q\).
Then we have \[ (a-22)^2 + (b-22)^2 = 2005 - 44^2/2 = 1037. \]
Note that 1037 has two and only two ways of being expressed as sum of two perfect squares: \[ 1037 = 14^2 + 29^2 = 19^2 + 26^2\]
WLOG, let \(0<a<b\), since \(a < 44 < b\) and \(a + b > r = 2005 - 44^2 = 69\), the only two solutions for \(0<a<b\) are \((a=36,b=51, ab=1836)\) and \((a=41,b=48,ab=1968)\). Hence the minimum value of \(ab\) is 1836.