Hilbert's third problem
From Academic Kids

The third on Hilbert's list of mathematical problems, presented in 1900, is the easiest one. The problem is related to the following question: given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces which can be reassembled to yield the second? Based on earlier writings by Gauss, Hilbert conjectured that this is not always possible. This was confirmed within the year by his student Max Dehn, who proved that the answer in general is "no" by producing a counterexample.
Note that the answer for the analogous question about polygons in 2 dimensions is "yes" and had been known for a long time; this is the BolyaiGerwien theorem.
Contents 
Dehn's answer
Dehn's proof is a nice instance of the general fact that algebra is often used to prove impossibility results in geometry. Other examples are squaring the circle and trisecting the angle.
We call two polyhedra scissorscongruent iff the first can be cut into finitely many polyhedral pieces which can be reassembled to yield the second. Obviously, any two scissorscongruent polyhedra have the same volume. Hilbert asks about the converse.
For every polyhedron P, Dehn defines a value, now known as the Dehn invariant D(P), with the following property:
 If P is cut into two polyhedral pieces P_{1} and P_{2} with one plane cut, then D(P) = D(P_{1}) + D(P_{2}).
From this it follows
 If P is cut into n polyhedral pieces P_{1},...,P_{n}, then D(P) = D(P_{1}) + ... + D(P_{n})
and in particular
 If two polyhedra are scissorscongruent, then they have the same Dehn invariant.
He then shows that every cube has Dehn invariant zero while every regular tetrahedron has nonzero Dehn invariant. This settles the matter.
A polyhedron's invariant is defined based on the lengths of its edges and the angles between its sides. Note that if a polyhedra is cut into two, some edges are cut into two, and the corresponding contributions to the Dehn invariants should therefore be additive in the edge lengths. Similarly, if a polyhedron is cut along an edge, the corresponding angle is cut into two. However, normally cutting a polyhedron introduces new edges and angles; we need to make sure that the contributions of these cancel out. The two angles introduced will always add up to π; we therefore define our Dehn invariant so that multiples of angles of π give a net contribution of zero.
All of the above requirements can be met if we define D(P) as an element of the tensor product of the real numbers R and the quotient space R/(Qπ) in which all rational multiples of π are zero. Both of these are vector spaces over Q and the tensor product is to be taken over Q as well.
Let l(e) be the length of the edge e and θ(e) be the dihedral angle between the two faces meeting at e, measured in radians. The Dehn invariant is then defined as
 <math>\operatorname{D}(P) = \sum_{e} l(e)\otimes (\theta(e)+\mathbb{Q}\pi)<math>
where the sum is taken over all edges e of the polyhedron P.
Further information
In light of Dehn's theorem above, one might ask "which polyhedra are scissorscongruent"? The beautiful answer is due to Sydler (1965): two polyhedra are scissorscongruent if and only if they have the same volume and the same Dehn invariant. In 1990, Dupon and Sah provided a simpler proof of Sydler's result by reinterpreting it as a theorem about the homology of certain classical groups.
Debrunner showed in 1980 that the Dehn invariant of any polyhedron with which all of 3d space can be tiled is zero.
Original question
Hilbert's original question was actually a bit more complicated: given any two tetrahedra T_{1} and T_{2} with equal base area and equal height (and therefore equal volume), is it always possible to find a finite number of tetrahedra, so that when these tetrahedra are glued in some way to T_{1} and also glued to T_{2}, the resulting polyhedra are scissorscongruent? Dehn's invariant can be used to yield a negative answer even to this stronger question.
History and motivation
The formula for the volume of a pyramid,
 1/3×base area×height,
had been known to Euclid, but all proofs of it involve some form of limiting process or calculus. Similar formulas in plane geometry can be proven with more elementary means. Gauss regretted this defect in two of his letters. This was the motivation for Hilbert: is it possible to prove the equality of volume using elementary "cutandglue" methods? Because if not, then an elementary proof of Euclid's result is also impossible.
References
 Max Dehn: "Über den Rauminhalt", Mathematische Annalen 55 (1902), pages 465478
 Sydler, J.P. "Conditions nécessaires et suffisantes pour l'équivalence des polyèdres de l'espace euclidean à trois dimensions", Comment. Math. Helv. 40 (1965), pages 4380
 Johan Dupont and ChihHan Sah: "Homology of Euclidean groups of motions made discrete and Euclidean scissors congruences", Acta Math. 164 (1990), no. 12, pages 127
 Hans E. Debrunner: "Über Zerlegungsgleichheit von Pflasterpolyedern mit Würfeln", Arch. Math. (Basel) 35 (1980), no. 6, pages 583587zh:希爾伯特第三問題