How many such equations?

Algebra Level 3

What is the cardinality of the set of equations of the form

$\left( x^2+ax+b \right)^{\left( x^2+cx+d \right)} = 1$

with integer coefficients $a,b,c,d$ such that all of their real solutions are integers?

countably infinite uncountably infinite finite

1 solution

K T
Jan 6, 2019

For shortness I will put $p_1=x^2+ax+b$ and $p_2=x^2+cx+d$ .

If $x$ is real, then $p_1$ and $p_2$ will be real too. To satisfy $p_1^{p_2}=1$ , we need either $p_1=1$ , or $p2=0$ , or $p_1=-1$ and $p2$ is an even integer.

Observe:

The set of integer n-tuples is countably infinite. This is taken as a known fact here.
If there are NO real solutions at all, then we can logically say that all of them are integers.
A subset of a countable set must itself be countable too.

So we only have to show that there are an infinite number of choices among the quadruples (a,b,c,d) for which there is no real solution for x. Take $a=0, b>1,c=0, d>0$ then $p_1>1$ and $p_2>0$ . For those choces $p_1^{p_2}=1$ has no real solutions. There are infinitely many such (a,b,c,d), so our final conclusion is:

There is a countably infinite number of quadruples (a,b,c,d) for which every real solution for x of $(x^2+ax+b)^{(x^2+cx+d)}=1$ is an integer solution.

There is also a countable infinite set of choices with actual integer solutions, namely choose $a=c=d=0, b>1$ , then $p_1^{p_2}= (x^2+b)^{x^2}$ which is a positive number to a nonnegative power. The only real solution is x=0.

K T - 2 years, 5 months ago

How many such equations?

1 solution

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