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Wednesday, November 16, 2022

Nested square roots

Let f(a) be the number of positive integers x such that x+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=a2a, x+x+a=a is an integer for positive integer a and thus f(a)1 for positive integer a.

Now suppose x+x+a=p

Some algebra gives us that 

x2(2p2+1)x+(p4a)=0

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

And so

(2p2+1)24(p4a)=q2

Which leads to

4a+1=(q2p)(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.

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