Reverse Engineer This Transcendental Equation!

Algebra Level 5

Let $a$ be any natural number, and let $f(x)={ a }^{ 2 }{ x }^{ 2 }-x-{ 1 }$ . Consider

${ f(x) }^{ { f(x) }^{ f(x) } }= { 2015 }$ .

If the sum of all solutions for $x$ is $k$ , determine $\frac { { \pi }^{ 2 } }{ k }$ . (This means if for some value of $x$ you can set a value of $a$ such that it satisfies the equation, then that value of $x$ is a solution)

The answer is 6.

1 solution

Dylan Pentland
Mar 3, 2015

Notice that for a unique $q$ , ${ q }^{ { q }^{ q } }=2015$ . Thus, for each value of $a$ , we wish to solve:

$a^{ 2 }{ x }^{ 2 }-x-1=q$

The sum of the roots for this is $\frac { 1 }{ { a }^{ 2 } }$ , for each value of $a$ . $q$ doesn't even matter! We now have that

$k=\sum _{ a=1 }^{ \infty }{ \frac { 1 }{ { a }^{ 2 } } } =\frac { { \pi }^{ 2 } }{ 6 }$ .

This is a pretty well-known result, so I won't include a proof here. (I'm pretty sure there's a wikipedia page if you're curious) Now that we have $k$ , the answer is $6$ .

Here is a proof , by a genius, in our Community itself , why go elsewhere ?

A Former Brilliant Member - 6 years, 3 months ago

in this question . why $q$ doesn't even matter ? @Dylan Pentland

Aman Rajput - 5 years, 10 months ago

The sum of the roots of a quadratic is independent of the constant term.

Dylan Pentland - 5 years, 10 months ago

I know ,,, but these powers of $q^{q^q}$ make me confused. ! -_-

Aman Rajput - 5 years, 10 months ago

Reverse Engineer This Transcendental Equation!

The answer is 6.

1 solution

0 pending reports