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.
No comments:
Post a Comment