Where are the integers?

Algebra Level pending

y = x*cosh(ln(x))

If x and y are both integers, find all possible values of x in terms of any integer n.

x = 1

x = 2^n - 1, n>0

x = \frac{n^2 + 1}{2}

x = n^2, n≠0

x = sinh(\frac{e^n}{n!})

x = 2\vert n\vert+1

x = \frac{(2n+1)^2 + 1}{2}

x = 2n+1

1 solution

A A
Sep 30, 2015

$cosh(x) = \frac{e^x + e^{-x}}{2}$

$cosh(ln(x)) = \frac{e^{lnx}+e^{-lnx}}{2} = \frac{x + 1/x}{2}$

The above still holds for negative values of x, because ln(-k) is defined in the complex numbers.

$y = x*cosh(ln(x))) = \frac{x^2 + 1}{2}, x≠0$

$2y = x^2 + 1, x≠0$

2y must be even, so $x^2+1$ must be even. Therefore, $x^2$ must be odd, and x must be odd as well. In other words, $\boxed{x = 2n+1}$ .