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=a2−a, √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+(p4−a)=0
For this to have an integer solution we need the discriminant to be a perfect square.
And so
(2p2+1)2−4(p4−a)=q2
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.
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