This is a problem in number theory, not quite a Diophantine equation; an IMO level one, so not very simple.
It can be transformed into a Diophantine equation:

isolve( x^2+y^2 = k*(1+x*y));

          {k = _Z1^2, x = _Z1, y = 0}

Actually this is given by Maple 2018 (on an old Laptop of mine). This is not correct without x>=0, y>=0,  for example x=2, y=-1, k=-5.
In Maple 2024 the result is NULL (also wrong, of course).

Note that isolve is not even able to solve directly the equation for k=4,  although it is of a Pell type. But we can help here:

isolve(z^2-3*y^2 - 4 = 0);  # x = 2*y + z

  ==>  z= ..., y=...

So, unfortunately isolve is not useful for our problem. The only thing we can do is to check numerically the result:

for i to 1000 do for j to i-1 do
  if irem(  ( i^2+j^2) , 1+i*j)=0 then lprint([i,j],(i^2+j^2)/(1+i*j)) fi
od od:

[8, 2], 4
[27, 3], 9
[30, 8], 4
[64, 4], 16
[112, 30], 4
[125, 5], 25
[216, 6], 36
[240, 27], 9
[343, 7], 49
[418, 112], 4
[512, 8], 64
[729, 9], 81
[1000, 10], 100

@mmcdara You have chosen the HermiteH polynomials in the Galerkin method. They are orthogonal in the interval (-oo, oo) with the weight exp(-x^2) you did not use.
So, why HermiteH? Probably ChebyshevT with the corresponding weight are better.

Most  coulditbe(f(z)=0) assuming real;  for not very simple f(z) are anyway (and inevitable) FAIL.

@acer  If we must start (for some reason) with Digits = 15, then method=_d01ajc  will not work, and the nonNAG methods are really slow. 

@sija  Your problem is to solve equations of the form  Int( f(t), t=-4 .. x) = a, f(t) and a being given.

The function F(x) = Int( f(t), t=-4 .. x)  is the solution of the ode  F'(t) = f(t), F(-4) = 0. So, F can be obtained by dsolve (numerically very efficient).

Finally, fsolve(F - a, -10..10)  finds the wanted approximate solution (in the interval -10..10)).
 

@Exiu 

convert(f, radical);

@Christian Wolinski 

coulditbe(exp(z)=0);        #? true
coulditbe(abs(z)+2+z=0);    #? true
solve(abs(z)+2+z=0);        #ok NULL

f can be represented this way iff 0 is not a pole for f (i.e. 0 is not a root of denom(f)).

@Andiguys  I have already mentionrd that Maximize does not accept double inequalities a <= b <=c; use a<=b, b <= c.
So, correct C2:=....

@Andiguys  You do not have a constraint plane. Using
solve(C1 union C3 union C3);
it results
dom := tau1=0..lambda, lambda= 0.0224489798..1.022448980;

Then:
plot3d(TRC(tau1, lambda), dom, orientation=[165, 75,0]);

@Andiguys  Use:

plot3d(TRC(tau1, lambda), tau1 = 0 ..1, lambda = 0 .. 1, orientation=[165, 75,0]);

@Andiguys Replace the execution group C3c := ...   with

temp := (-2*R0er*d^2*lambda + 2*d^2*g1*lambda + Aer*d - Am*d + 2*Clr*lambda + Rer*d - Rm*d)/(2*(-R0er*d^2 - R0m*d^2 + d^2*g1 + d*g2 + Clm + Clr));
C3c := {0 <= temp, temp <= 1};

Use the same intervals for tau1 and lambda in Minimize and plot!

@ecterrab Please note that the formula is valid for all z in C, not just |z|<Pi/2 (unlike the power series one).

@ecterrab  The expansion comes from Mittag-Leffler theorem  Mittag-Leffler's theorem - Wikipedia.
The singularities are exactly those of tan, so your objection is not justified.
But of course, FunctionAdvisor cannot list all the possible formulae.
 

@nm This is obviously a bug. Actually a regression bug, inexistent in Maple 2023 (and an elementary one, the expression being continuous w.r.t. x).
Unfortunately, today is much more difficult to find an explanation than 20 (or even 10)  years ago!