A cute problem from New Zealand Maths Olympiad

 

In the image above, ABCD is a square, and BX = DY, X lying on BC, and Y on CD extended.

P is the intersection point of the diagonal BD and XY. 

Show that PY = PX. 

(Diagram not to scale!)

Try using pure geometric methods only.

Solution diagram below:

Nested square roots

Let $f(a)$ be the number of positive integers $x$ such that $$\sqrt{x + \sqrt{x + a}}$$ is a positive integer. 

 Show that there are infinitely many positive integers $a$ such that $f(a) = 1$

Scroll down for a solution.

Note that for $x = a^2 - a$, $\sqrt{x + \sqrt{x+a}} = a$ is an integer for positive integer $a$ and thus $f(a) \ge 1$ for positive integer $a$.

Now suppose $$\sqrt{x + \sqrt{x+a}} = p$$

Some algebra gives us that 

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

For this to have an integer solution we need the discriminant to be a perfect square. 

And so

$$(2p^2 + 1)^2 - 4(p^4 - a) = q^2$$

Which leads to

$$4a +1 = (q-2p)(q+2p)$$

Thus if $4a+1$ is prime, there is a unique solution.

Since there are infinitely many primes of the form $4a+1$, we are done.