Tricky positive diophantine equation Find all positive integer solutions to the equation $x^3=y^5+100$.
Notice that $7^3=3^5+100$ is a solution, but I can't find any more.
Thanks for your help!
P.S. I don't know any advanced number theory, just basic olympiad stuff.
 A: Again not a complete solution, just some new insights...
The fact that you already know a solution makes it interesting. It allows us to rewrite the equation as
$$x^3-7^3=y^5-3^5,$$
or
$$(x-7)(x^2+7x+49)=(y-3)(y^4+3y^3+9y^2+27y+81).$$
Note that every prime divisor of $\frac{x^3-7^3}{x-7}$ is either $3$ or congruent to $1$ modulo $3$. Similarly, every prime divisor of $\frac{y^5-3^5}{y-3}$ is either $5$ or congruent to $1$ modulo $5$. (This is all due to a more general result involving cyclotomic polynomials of primes.)
The factor that 'makes' the RHS large is $y^4+3y^3+9y^2+27y+81$. The fact that this can only have primes $\equiv0,1\pmod5$ suggests that there will not be many solutions as $x^3-7^3$ might have a lot of prime divisors that have to divide $y-3$. Finding a close upper bound on the primes dividing $x^3-7^3$ that are $\equiv0,1\pmod5$ possibly will exclude further solutions.
I'll start off considering an easy case.  If $x-7\mid\frac{y^5-3^5}{y-3}$ then either $x\equiv2$ or $x\equiv3\pmod5$. In any case $x^2+7x+49$ is not congruent to $0$ or $1$ modulo $5$, so it should have some divisor $>1$ in common with $y-3$. We want this divisor to be big in order to obtain a contradiction.
Perhaps finding an upper bound on $\gcd(x-7,y-3)$ might help. Certainly the $\gcd$ should divide $3x-7y$ but I can't really make anything interesting of this.
A: Thanks to Edward we are morally certain that $(x,y) = (7,3)$
is the only solution.  [I extended the search to all odd $y<10^9$ using
the gp command
forstep(y=1,10^9,2, if(ispower(y^5+100,3),print(y)))

which took about 4 minutes and output only $3$ as expected.]
Also, as a special case of

Siegel's theorem on integral points we know that
there are only finitely many integer solutions.
The original proof is "ineffective", i.e. doesn't yield
an algorithm that can be guaranteed to find all solutions;
more recent work sometimes provides such algorithms, and even ones
that can be carried out in practice, but they're not easy, being
clearly "advanced number theory" as opposed to "basic olympiad stuff".
The existence of one nontrivial solution $(7,3)$ makes it quite unlikely
that an elementary proof is possible; for starters it will never be enough
to just use congruences modulo a few small numbers, because once $(7,3)$
works there will always be infinitely many more surviving candidates.
One thing that makes this just a bit more tractable,
and effectively solvable at least in principle, is that $100$
happens to be a square, and also cube-free, so any solution yields
a pair of solutions $(\pm 10,-x,y)$ to the Diophantine equation
$X^2+Y^3+Z^5=0$ in relatively prime integers.  This lets us build on
previous work in the paper

J. Edwards: A Complete Solution to $X^2 + Y^3 + Z^5 = 0$,
  Journal f. d. reine und angew. Math. (Crelle's Journal)
  571 (2004), 213-236.

which completely solves this equation under the relatively-prime condition.
Edwards finds $27$ identities $X_i(r,s)^2 + Y_i(r,s)^3 + Z_i(r,s)^5 = 0$,
each with $X_i,Y_i,Z_i$ homogeneous polynomials of degree $30$, $20$, $12$
respectively, and shows that every solution of $X^2+Y^3+Z^5=0$
in relatively prime integers is obtained from one of these solutions
by substituting some integers for $r$ and $s$.  Thus any solution with
$X = \pm 10$ is a solution of one of $27$ pairs of
Thue equations
$X_i(r,s) = \pm 10$.  This count is probably exaggerated because
some $i$ might be ruled out by congruence conditions, but at least
one must be possible, again because we know the solution $(X,Y,Z)=(10,-7,3)$.
Unless we're quite lucky, we won't be able to solve all these equations
by elementary means, and since each of the polynomials $X_i(r,s)$ 
has degree $30$ the more advanced techniques (try $p$-adic methods first,
and the Baker bounds where the $p$-adic analysis does not reach a
complete solutions) could take quite some work to complete.
[added later] The sixth of J.Edwards' identities has 
$Z = \sum_{j=0}^{12} a_j r^j s^{12-j}$, 
where the coefficients $a_0,\ldots,a_{12}$ are
$$
3,12,-132,0,-1980,-3168,3168,12672,-39600,-10560,-61248,-26112,27072,
$$
and then (as with all these identities) $Y$ is the scaled Hessian
$(Z_{rr} Z_{ss} - Z_{rs}^2)/132^2$ and $X$ is the scaled Jacobian
$(Y_r Z_s - Y_s Z_r) / 240$.  Taking $(r,s)=(0,1)$ recovers the solution
$10^2 - 7^3 + 3^5 = 0$.  So we must at least prove that there are
no other solutions of the degree-$30$ Thue equation $Z_6(r,s) = 10$.
Since $Z_6$ is irreducible and is not positive-definite (the polynomial
$Z_6(1,s)$ has four real roots), even this part of the problem 
seems unlikely to have an elementary solution.
