IMO-1959

1. Prove that the fraction \(\frac{21n+4}{14n+3}\) is irreducible for every natural number \(n\).
• To show that a fraction is irreducible, we need to prove that the numerator and denominator are relatively prime. Note that we have \(3 (14n+3) - 2 (21n+4) = 1\), \(\forall n \in \mathbb{N}\). Let \(d\) be the g.c.d of \(21n+4\) and \(14n+3\). Hence, \(d | \left( 3 (14n+3) - 2 (21n+4) = 1 \right) \). Hence, \(d | 1 \implies d = 1\). Hence, \(21n+4\) and \(14n+3\) are relatively prime. Hence, the fraction \(\frac{21n+4}{14n+3}\) is irreducible for every natural number \(n\).
2. For what real values of \(x\) is \[\sqrt{\left( x + \sqrt{2x-1} \right)} + \sqrt{\left( x - \sqrt{2x-1} \right)} = A,\] given (a) \(A = \sqrt{2} \), (b) \(A = 1\), (c) \(A = 2\), where only non-negative real numbers are admitted for square roots?
• Only non-negative real numbers are admitted for square roots, we need \(2x-1 \geq 0\) and \(x \geq \pm \sqrt{2x-1}\). The two conditions yield us \(x \geq \frac12\). Squaring both sides, we get, \[\left( x + \sqrt{2x-1} \right) + \left( x - \sqrt{2x-1} \right) + 2 \sqrt{\left( x + \sqrt{2x-1} \right)\left( x - \sqrt{2x-1} \right)} =A^2\] \[2x + 2 \sqrt{x^2 -2x + 1} = A^2\] \[2x + 2 \left \lvert x - 1 \right \rvert = A^2 \]
• If \(x \geq 1\), we get \[2x + 2x - 2 = A^2 \implies 4x = A^2 + 2 \implies x = \frac{A^2 + 2}{4}\] Hence, if \(A \geq \sqrt{2}\), we have \(x = \frac{A^2 + 2}{4}\).
• If \(\frac12 \leq x \leq 1\), we get \(2x + 2 - 2x = A^2 \implies 2 = A^2\), \(\forall x \in \left[\frac12,1\right]\).
• So the solution can be written as follows: \(x = \left \{ \begin{array}{lr} \frac{A^2+2}{4} & \text{If } A \gt \sqrt{2}\\ \text{Any value }\in \left[\frac12,1\right] & \text{If } A = \sqrt{2} \\ \text{No solution } & \text{If }A \lt \sqrt{2} \end{array} \right. \)
• For the current problem, we get \(x = \left \{ \begin{array}{lr} \frac{3}{2} & \text{If } A = 2 \\ \text{Any value }\in \left[\frac12,1\right] & \text{If } A = \sqrt{2} \\ \text{No solution } & \text{If }A = 1 \end{array} \right. \)
3. Let \(a,b,c\) be real numbers. Consider the quadratic equation in \(\cos(x)\) : \[ a \cos^2(x) + b \cos(x) + c = 0. \] Using the numbers \(a,b,c \) form a quadratic equation in \(\cos(2x)\), whose roots are the same as those of the original equation. Compare the equations in \(\cos(x)\) and \(\cos(2x)\) for \(a=4\), \(b=2\) and \(c=-1\).
• We have \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\). Hence, the equation becomes \[\displaystyle a \left( \frac{1 + \cos(2x)}{2} \right) + b \sqrt{\frac{1 + \cos(2x)}{2}} + c = 0\] Rearranging and squaring both sides, we get \[\left( a \left( \frac{1 + \cos(2x)}{2} \right) + c \right)^2 = b^2 \left( \frac{1 + \cos(2x)}{2} \right) \] \[ a^2 + a^2 \cos^2(2x) + 4c^2 + 2a^2 \cos(2x) + 4ac + 4ac \cos(2x) = 2b^2 + 2b^2 \cos(2x)\] \[a^2 \cos^2(2x) + (2a^2 + 4ac -2 b^2) \cos(2x) + (a^2 + 4c^2 + 4ac - 2b^2) = 0\] Plugging in \(a = 4, b=2\) and \(c=-1\), we get the equation in terms of \(\cos(x)\) as \[4 \cos^2(x) + 2 \cos(x) - 1 = 0\] and the equation in terms of \(\cos(2x)\) as \[16 \cos^2(2x) + 8 \cos(2x) - 4 = 0.\] Hence, \(\cos(x)\) and \(\cos(2x)\) satisfy the same quadratic equation.