Has the sum of 4 cubes problem been proven? Today in class, my professor was lecturing on the sum of 3 cubes and sum of 4 cubes problems. Namely, can every number be written as the sum of 3 (or 4) cubes? He discussed their origins and showed a few examples, and showed how difficult they could be to find for certain numbers (such as 33 or 42 for the sum of 3 cubes). He said we would not cover their proofs in the course because they were "beyond the scope of the course." When I went to look them up, however, it seems as though they are open problems and have not been proved. I don't think my professor would get this wrong, so I'm a bit confused. I would appreciate any clarification. If they have been proven, where can I see the proofs?

First:You can use this identity for finding numbers which are the sum of three cubes:
$$(x-y)^3+(y-z)^3+(z-x)^3=3(x-y)(y-z)(z-x)$$
For example:
$(3-5)^3+(5-7)^3+(7-3)^3=3(3-5)(5-7)(7-3)= 48$
Second : we solve this problem to find a number which it's cube is the sum of three cubes:
$$x^3+y^3+z^3=u^3$$
Let $u=-t$ we have:
$$x^3+y^3+z^3+t^3=0\space\space\space\space(1)$$
This equation has infinitely many solutions(positive or negative), as you will see they make a set of particular numbers which means not every cube can be written as the sum of three cubes.
Suppose $a, b, c, d , \alpha, \beta, \gamma, \delta $ are two groups of four numbers that satisfy equation (1) . Choose $k$ such that numbers $a+k\alpha, b+k\beta, c+k\gamma, d+k\delta$ also satisfy equation (1), or we can have:
$$(a+k\alpha)^3+(b+k\beta)^3+(c+k\gamma)^3+(d+k\delta)^3=0$$
we expand each term; considering groups (a, b, c , d ) and $(\alpha, \beta, \gamma, \delta)$ both satisfy the equation i.e.:
$a^3+b^3+c^3+d^3=0$
$\alpha^3+ \beta^3+ \gamma^3+ \delta^3=0$
We have:
$3a^2k\alpha+3ak^2\alpha^2+3b^2k\beta+3bk^2\beta^2+3c^2k\gamma+\3ck^2\gamma^2+3d^2k\delta+3dk^2\delta^2=0$
Or:
$3k[(a^2\alpha+b^2\beta +c^2\gamma+d^2\delta)+k(a\alpha^2+b\beta^2+c\gamma^2+d\delta^2)]=0$
This relation can be zero if one of it's factors is zero. Equating each factor to zero gives two values for k; one is $k=0$(which is not of our interest because it means we do not add anything to numbers a, b, c and d), second is:
$k=-\frac{a^2\alpha +b^2\beta+c^2\gamma+d^2\delta}{a\alpha^2+b\beta^2+c\gamma^2+d\delta^2}\space\space\space (2)$
If we have two groups of solutions we can find a new group of four numbers as the solution to equation (1).For this we k times of numbers of first group to numbers of second group , provided k is found by relation (2). To do this we need to have a group of solutions, say $(x, y, z, t)=(3, 4, 5, -6). To find second group let:
$\alpha, \beta, \gamma, \delta)=(r, -r, s, -s)$
clearly these numbers satisfy equation (1). Putting these values in (2) we get:
$$k=\frac{7r+11s}{7r^2-s^2}$$
So we have:
$a+k\alpha=\frac{28r^2+11rs-3s^2}{7r^2-s^2}$
$b+k\beta=\frac{21r^2-11rs-4s^2}{7r^2-s^2}$
$c+k\gamma=\frac{35r^2+7rs+6s^2}{7r^2-s^2}$
$d+k\delta=\frac{-42r^2-7rs-5s^2}{7r^2}$
In this way general form of solutions, considerin all numerators are equal, can be:
$x=28r^2+11rs-3s^2$
$y=21r^2-11rs-4s^2$
$z=35r^2+7rs+6s^2$
$t=-42r^2-7rs-5s^2$
For example take $r=s=1$ you get:
$(x, y, z, t)=(1, 6, 8, 9)$
$1^3+6^3+8^3=9^3$
