@Markiyan Hirnyk  As you can see, it took a lot of options for finding all solutions. Here's an easier way. From continuity and properties of functions it follows that two roots are in the ranges  x=-1..0  and  x=1..2  and third exact root  x=100  is evident.

Eq:=100^x=x^100:

fsolve(Eq, x=-1..0);

fsolve(Eq, x=1..2);

                  -0.9568903883

                    1.049519190

Another way:

RootFinding[Analytic](100^x=x^100, re=-1..110, im=-1..1):

select(t->type(t, realcons), [%]);

                  [-0.956890388296380, 1.04951918980717, 100.000000000000]

@Carl Love  Thank you, but I have already made the amendment earlier.

@Carl Love  That's good, but it works only for newer versions Maple. Variant for all versions:

P:=a*x^2*y*z+b*x*y*z+c*x^2*y*z+d*x*y^2*z+e*x*y*z:

L:=[coeffs(P, [x,y,z], 't')]: 

H:= table(zip(`=`, [t], L)):

H[x^2*y*z];

                         a+c

@Axel Vogt  Yes, this product is the classical identity. See here

@Axel Vogt  Thank you for your interest and useful procedure. I think this procedure solves and other similar examples. There are also examples with which it can not handle.

Examples:

f:=n->tan(sqrt(n^2+2*n/3)*Pi);

seqLimit(f(n), n, 1);

h:=n->product(cos(1/2^k), k=1..n);

evalf(limit(h(n), n=infinity));

identify(%);

seqLimit(h(n), n);

@Carl Love  Actually, this sequence increases monotonically. This follows from the monotony of  sin  function on [0,Pi/2] and

g:= n->sqrt(n^2+n)*Pi - Pi*n:

is(diff(g(n), n)>0)  assuming n>1;

is(g(n)<Pi/2)  assuming n>1;

                             true
                             true

@Carl Love   Thanks.

Here is another interesting phenomenon. With an increase  n  the calculation accuracy falls:

f:=n->sin(sqrt(n^2+n)*Pi)^2;

seq(evalf(f(10^5*k)), k=1..10);

@Kitonum   We can solve the problem numerically by reducing it to the calculation of the sum a series:

f:=n->sin(sqrt(n^2+n)*Pi)^2;

evalf[100](sum(f(n+1) - f(n), n=1..infinity)+f(1));

@Carl Love  I think your solution closer to hand than to automatic. Why we have to guess to use asympt(sqrt(n^2+n)*Pi, n, 1) ?  We wish the developers to solve this problem in a future release of Maple.

BTW in Mathematica the problem already solved:

Limit[Sin[Sqrt[n^2 + n]*Pi]^2, n -> Infinity, Assumptions -> Element[n, Integers]]

                                                                    1

@tomleslie  Thanks. And how about this example for  n::posint  in Maple:

limit(sin(sqrt(n^2+n)*Pi)^2, n=infinity);

@Markiyan Hirnyk  Why did you write the option "solutions=16" ? Without this option, DirectSearch finds some the other number of solutions. I think that aser's and Preben's ways (using only Maple tools) are more  justified.

@Carl Love   Of course, you're right! Still, the list is somewhat broader. As it is written in the help, for example, the list includes the values of functions with rational arguments and algebraic numbers of degree <=6.

Examples:

Digits:=20:

evalf(ln(101));

identify(%);

convert(%, ln);

evalf(sqrt(3+5^(1/3)));

identify(%);

evalf(sin(1/3));

identify(%);

Markiyan Hirnyk  Please provide a link to the source of this example. It is very similar to the challenges that are offered on student competitions.

@Markiyan Hirnyk  In fact squaring - this is a very natural idea, which is often useful. Here is another solution without complex numbers and conversions:

expr:=expand(sin(-(1/6)*Pi+(1/2)*arccos(1/3)));

z:=1/2*arccos(1/3):  # Obviously  cos(z)>0  and  sin(z)>0

eval(expr, [cos(z)=sqrt(combine(cos(z)^2)), sin(z)=sqrt(combine(sin(z)^2))]);

@Markiyan Hirnyk   The ratio of axes of the ellipse is equal to

sqrt(160*(1/4))/sqrt(160/9);

                        3/2