A usual suspect for seeking solutions to this type of differential equations
is any linear combination of functions of the form e^{zx}.

Taking the tenth derivative leads to *z*^{10}e^{zx},
so we are looking for a *z* value that is a tenth root of unity:
*z*^{10}=1. This has 10 solutions, of which some are primitive
roots, so satisfy the condition that lower order derivatives will not equal
the original function.

Notably, the conjugate of a primitive root is itself a primitive root, so by adding the two together we still get a function that satisfies all the criteria listed so far, but is, additionally, also a real function.

Up to a factor of 2, this can simply be stated as Re(e^{zx}),
and *z* can be any primitive tenth root of unity, such as, for example,
e^{iπ/5}=cos(π/5)+*i*sin(π/5).

So, now, it just becomes a question of wording the function in a way that makes no mention of the intermediate complex values:

Re(e^{zx})=Re(e^{Re(z)x}(cos(Im(*z*)*x*)+*i*sin(Im(*z*)*x*))=e^{Re(z)x}cos(Im(*z*)*x*)

Leading to one possible *f* being
e^{cos(π/5)x}cos(sin(π/5)*x*).

This is enough to solve the problem but not the bonus question, because π is not computable on the target pocket calculator.

Luckily, it is possible to calculate both cos(π/5) and sin(π/5):
cos(π/5)=(1+sqrt(5))/4 and sin(π/5)=sqrt(1-cos^{2}(π/5)).

Here are two ways how to prove these values. The first, purely algebraic, comes from Andreas Stiller:

For every *x*
cos(2*x*)=2cos^{2}(*x*)-1.

cos(π/5)=cos(π-4π/5)=-cos(4π/5).

Substituting in the two connections we have between cos(π/5), which we will
call *c* and cos(4π/5) we get:

2(2*c*^{2}-1)^{2}-1=-*c*

so

8*c*^{4}-8*c*^{2}+*c*+1 =
(*c*+1)(2*c*-1)(4*c*^{2}-2*c*-1) = 0.

This equation has four solutions, but only one of them is in the search range
between cos(π/4)=sqrt(2)/2 and cos(π/6)=sqrt(3)/2. It is
*c*=(1+sqrt(5))/4.

A completely different take on this proof was sent in by JJ Rabeyrin. JJ's approach is geometric:

Consider triangle *ABC*, isosceles in *A*, where:

*BC*= 1- angle
*CBA*= 2π/5 - angle
*BCA*= 2π/5 - and thus angle
*BAC*= π/5

Let *D* be the intersection of *AC* and the bisector of angle *ABC*. We see that:

- angle
*DBA*= angle*CBA*/2 = π/5 - angle
*BAD*= angle*BAC*= π/5 - thus
*DBA*is isosceles in*D* - and thus
*DB*=*AD*

Moreover, we can observe that *BCD* is isosceles in *B*, so
*BD* = *BC* = 1 = *AD*.

Looking at *DAB*, we have:

and thus *AB* = 2cos(π/5).

Looking at *DBC*:

So, *CD* = 2cos(2π/5).

Because *ABC* is isosceles in *A*, we now have *AB* = *AC* = *AD* + *DC*, leading to

We can now integrate this geometric insight with

I will note that to actually calculate the required function with the
designated pocket calculator does require *x* to be input twice.

Thank you, again, to all readers who have followed Using your Head is Permitted throughout these 10 years, and thank you to all readers who joined along the way.

And, of course, a big thank you to Ori Pomerantz for the anniversary riddle.