Hello, $(4159, 3301, 2620)$ does nicely. To motivate this solution, for integers $a,b,c$ define $N(a + b\sqrt[3]{2} + c \sqrt[3]{4}) = a^3 + 2b^3 + 4c^3 - 6abc$. It is not terribly hard to check that this function is multiplicative (this expression is found by computing $(a + b\sqrt[3]{2} + c \sqrt[3]{4})(a + b\omega \sqrt[3]{2} + c \omega^2 \sqrt[3]{4})(a + b \omega^2 \sqrt[3]{2} + c \omega \sqrt[3]{4})$ which is motivated by automorphisms). Now, remark that as $1 + \sqrt[3]{2} + \sqrt[3]{4}$ is a unit in $\mathbb{Z}[\sqrt[3]{2}]$ (its inverse is $\sqrt[3]{2} - 1$), it follows its norm is $1$. Now just raise $1 + \sqrt[3]{2} + \sqrt[3]{4}$ to a large power to get components all over $2010$. Darn this is an incredibly silly problem. pythag011 was right that there was an incredibly silly TST problem that was killed by looking at norms in a certain domain.

@dinoboy: hey, look, I don't get your solution. I've heard the word "norm", but I have no idea, how does this work. Do you have some documents about it?

You could also note that if $(a,b,c)$ is a solution, so is $(a+2b+2c,a+2c+b,a+b+c)$, which is essentially the same idea.

polya78 wrote: You could also note that if $(a,b,c)$ is a solution, so is $(a+2b+2c,a+2c+b,a+b+c)$, which is essentially the same idea. Hooray for revivals. What exactly is your motivation for transforming $(a, b, c)$ to $(a + 2b + 2c, a + b + 2c, a + b + c)$?

v_Enhance wrote: Determine, with proof, whether or not there exist integers $a,b,c>2010$ satisfying the equation \[a^3+2b^3+4c^3=6abc+1.\] This problem has been on AoPS for long, in fact it asked to prove that there dont exist such $a,b,c$ I dont remember, but I suppose it was djmathman who had posted it? It has also been on the blog of some user, it was the same avatar as that of djmathman... But I cant find it. Can anyone help?

The substitution follows from the fact that if $P(x,y,z)=x^3+y^3+z^3-3xyz$, and if $w^3=2$, and $x'=x+w^2y+wz,y'=wx+y+w^2z,z'=w^2x+wy+z$, then $P(x',y',z')=P(x,y,z)$. To prove this is so, note that $P(x,y,z)$ is the determinant of $\begin{pmatrix} x & y & z \\ z & x & y \\ y & z & x \end{pmatrix}$, which means that $P(x',y',z')$ is the determinant of $\begin{pmatrix} x & y & z \\ z & x & y \\ y & z & x \end{pmatrix}* \begin{pmatrix} 1 & w & w^2 \\ w^2 & 1 & w \\ w & w^2 & 1 \end{pmatrix}$

We are motivated by the identity $ \left|\begin{matrix}a & \sqrt[3]{2}b & \sqrt[3]{4}c\\ \sqrt[3]{4}c & a & \sqrt[3]{2}b\\ \sqrt[3]{2}b & \sqrt[3]{4}c & a\end{matrix}\right| = a^3 + 2b^3 + 4b^3 - 6abc $. Since the triple $ (1, 1, 1) $ works we can plug this in for $ (a, b, c) $ and raise the matrix to arbitrarily high powers to obtain the desired result.

Sorry to revive, but does anyone here have a good books / paper about using arithmetic in a certain domain like what dinoboy did? I have found a good one i.e Arithmetic extensions of Q by Dusan Djukic. Are there any other books? Thank you

MrRTI wrote: Sorry to revive, but does anyone here have a good books / paper about using arithmetic in a certain domain like what dinoboy did? There are a few chapters on this in my current draft of http://www.mit.edu/~evanchen/napkin.html.

OK, let me write out a very short summary, although without some commutative and linear algebra this might not be very comprehensible. A number field $K$ is a finite-dimensional field extension of $\mathbb Q$. Examples include $\mathbb Q(\sqrt 2)$ (which has basis $1$, $\sqrt 2$), $\mathbb Q(i)$ (which has basis $1$, $i$), and $\mathbb Q(e^{\frac{2\pi i}{5}})$ (which is five-dimensional with basis the fifth roots of unity) and so on. Let's say $K$ has degree $n$. Given an $\alpha \in K$, there are two equivalent ways to define the norm: 1. Take the $n$ embeddings $\sigma_1, \dots, \sigma_n$ of $K$ into $\mathbb C$, and multiply them together. The elementary phrasing is look at the "conjugates" of $\alpha$, and multiply them all together (including $\alpha$ itself). For example, in $\mathbb Q(i)$ the norm of $a+bi$ is $(a+bi)(a-bi) = a^2+b^2$. The catch is that you have to do this with "multiplicity": $2$ should have norm $2^2=4$, even though $2$ doesn't actually have conjugates. That's why the phrasing with embeddings is better. 2. Look at the map $K \to K$ by $x \mapsto \alpha x$. Take the determinant of this map. This is the norm. Example: in $\mathbb Q(i)$, if we take a basis $1$, $i$ then multiplying by $a+bi$ is a matrix $\left( \begin{array}{cc} a & -b \\ b & a \end{array} \right)$ which has determinant $a^2+b^2$. You can show these definitions are equivalent. The first definition tells you the IMO natural way to compute norms, and by Vieta's formulas you can see it's an integer. The second definition tells you that norms are multiplicative (the norm of a product is the product of the norms). This problem is just the special case where $K = \mathbb Q(\sqrt[3]{2})$, which has basis $1$, $\sqrt[3]{2}$, $\sqrt[3]{4}$. The norm of $a + b \sqrt[3]{2} + c \sqrt[3]{4}$ is $a^3+2b^3+4c^3-6abc$.

Can the methods in the above posts be generalized to find all the solutions to the original equation? (over positive integers, of course)

Yes, they can, since all we wish to do is find the units of $O_K$ where $K=\mathbb Q(\sqrt[3]{2})$, which are finitely generated and in fact we know how many generators we need because of https://en.wikipedia.org/wiki/Dirichlet%27s_unit_theorem.

Hello $(16001,12700,10080)$ is also a solution

There is a unique triple $(a,b,c)$ of two-digit positive integers $a,\,b,$ and $c$ that satisfy the equation $$a^3+3b^3+9c^3=9abc+1.$$Compute $a+b+c$

soy_un_chemisto wrote: polya78 wrote: You could also note that if $(a,b,c)$ is a solution, so is $(a+2b+2c,a+2c+b,a+b+c)$, which is essentially the same idea. Hooray for revivals. What exactly is your motivation for transforming $(a, b, c)$ to $(a + 2b + 2c, a + b + 2c, a + b + c)$? Think of it as follows - by trial and error find positive constants $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3$ such that if $(a, b, c)$ is a solution, then $(x_1a + y_1b + z_1c, x_2a + y_2b + z_2c, x_3a + y_3b + z_3c)$ also is. These constants must be very small in order for equating coefficients at $a$, $b$, $c$ in $$a^3 + 2b^3 + 4c^3 - 6abc = (x_1a + y_1b + z_1c)^3 + 2(x_2a + y_2b + z_2c)^3 + 4(x_3a + y_3b + z_3c)^3 - 6(x_1a + y_1b + z_1c)(x_2a + y_2b + z_2c)(x_3a + y_3b + z_3c)$$to work. In fact, the coefficient at $a^3$ is $x_1^3 + 2x_2^3 + 4x_3^3 - 6x_1x_2x_3 = 1$, so take $x_1=x_2=x_3=1$; the coefficient at $b^3$ is $y_1^3 + 2y_2^3 + 4y_3^3 - 6y_1y_2y_3 = 2$, so take $y_1 = 2$, $y_2 = 1$, $y_3 = 1$; the coefficient at $c^3$ is $z_1^3 + 2z_2^3 + 4z_3^3 - 6z_1z_2z_3 = 4$, so take $z_1 = 2$, $z_2 = 2$, $z_3 = 1$. Now opening the brackets shows that the required identity indeed holds.