Current Events > For which integers x is sqrt[x+sqrt(x+8)] a positive integer?

(+) Yeah! (+)

lderivedx posted...

I liked this one.

Answer:

We set sqrt(x+sqrt(x+8)) = a, for a some positive integer. Squaring out the roots then collecting coefficients, we get the equation:

0 = x^2 - (2a^2+1)x+(a^4-8)

Using the quadratic formula we notice that since (2a^2+1) is odd, (2a^2+1)^2-4(a^4-8) must be a perfect square and odd. The latter simplifies to 4a^2+33, so we set it equal to the square of an odd integer 2k+1 to get:

4a^2+33=(2k+1)^2

Subtract the (2k+1)^2 over and factor using difference of squares to get

(2a+2k+1)(2a-2k-1)+33=0

The product there must be an either an (11 and a 3) or a (33 and a 1), where one of them is negative. Solving this, the only positive solution to the former that works is when a=2 and a=8 for the latter. Going back to the polynomial in x^2, we solve and we get four solutions for x: 1, 8, 56, and 73. These are not guaranteed to solve the original equation, but all solutions to the equation are in that list, so you can check that the solutions are 1 and 56.

sick

Robot2600 posted...

surreals are actually real numbers

complex(imaginary), irrational, and transcendental numbers are not

transcendental numbers are also real numbers. they're the ones that aren't solutions to polynomials with integer coefficients.