Unachievable sums of cubes and powers (fewnomial Fermat bounds)

Question

I want to prove that there are an infinite number of integers that can't be written as the sum of two cubes .

I tried to consider integers in the form of $10^{3n+1}$ but I feel there is something wrong with this approach .

Answer 1

Suppose $n = a^3 + b^3$ for some integers $a, b$. Reducing this equation $\!\bmod 7\,$ shows that the only possible residues of $n$ are $0, \pm1,\pm2,\,$ since cubes have residue $0$ or $\pm1$ (as is easily verified).

Thus if $n \not\equiv 0,\pm1,\pm2\pmod{\! 7}\,$ it follows that $n$ is not representable as the sum of $2$ cubes. There are infinitely many such $n$, namely all $\,n\equiv \pm3\pmod{\!7}$.

Answer 2

It is worth explicitly emphasizing the obvious generalization of the method in fwd's answer - which is a common method of proving certain Diophantine equations unsolvable - namely by choosing a modulus where the summands take too $\rm\color{#0a0}{few\ values}$ to achieve all residues. For example, if we have a sum of $\,j\,$ terms that take only values $\equiv \pm1\,$ or $\,0\pmod{\!n}\,$ then their sum $\,s\,$ is congruent only to values in the range $\,-j\le s\le j,\,$ so if our modulus is $\,n =2k+1 > 2j+1\,$ then the sum cannot take the values $\pm \{\color{#0af}{j\!+\!1},j\!+\!2,\ldots,k\}\bmod n\,$ (using a balanced residue system).

A common example of a summand taking only the few values $\equiv \pm1,0\,$ is $\,y = x^{m(p-1)/2}\,$ in $\,\Bbb Z_p,\,$ since by $\rm\color{#c00}{Fermat\!\!:}\,$ if $\,x\not\equiv 0\,$ then $\,y^2\!\equiv \color{#c00}x^{\color{#c00}{(p-1)}m}\!\equiv\color{#c00}1\,$ thus $\,y\equiv \pm1.\,$ Clearly a monomial formed from any product of such $\:\!x_i^{m_i(p-1)/2}$ also takes said $\rm \color{#0a0}{few\ values}$. So by above, if prime $\,p=2k\!+\!1\,$ and our sum has $\rm\color{#0a0}{fewer}$ than $\:\!k\:\!$ such monomial summands then it cannot take all residues $\!\bmod p,\,$ e.g. $\,\color{#d0f}{11}\:\! x_1^{6} x_2^{12}\! +10 \:\! x_3^{18} x_4^{24}\! = \color{#0af}{6}\!+\!13 x_5\,$ is unsolvable in integers $\:\!x_i\,$ by $\!\bmod 13\!:\:\! \color{#d0f}{11},10\equiv \color{#d0f}{-2},-3\,$ and $\,|\color{#d0f}{{-}2}|+|{-}3| < \color{#0af}6\,$ (using balanced residue system $\pm\{0,1,2,\ldots,\color{#0af}6\}\bmod 13).\,$ [We can also compute the exact values taken: $\color{#d0f}{11}\{0,\pm1\}+10\{0,\pm1\}\equiv $ $\{0,\color{#90f}{\pm2}\}+\{0,\pm3\}\equiv $ $\pm\{0,1,2,3,5\}$ (so it omits $\,\pm 4\,$ too), but this is generally more work - which is not needed in cases like that above, where said (extreme) bounds suffice].

The above method works because the powers $\,x^{m(p-1)/2}\,$ have residues constrained to a $\rm\color{#0a0}{short}$ interval $[-1,1],\,$ being either $0$ or square roots of $\:\!1.\,$ A similar method may apply to modular $\:\!k$'th roots of $1$ when they lie in a short interval, e.g. the cube roots of $1$ modulo $73,\,$ are $\,-9,1,8\,$ so $\,x^{24}+y^{24}+z^{24}\,$ lies in interval $[-27,24]\,$ so it is $\,\not\equiv 25,26,27,\pm28,\pm29,\ldots,\pm36\pmod{\!73}$. Again, we could also exclude more residues by further computing the exact residues taken by a sum of $\:\!3\:\!$ residues in $\{-9,1,8\}\ (= $ cube roots of one).

This method (and minor variants) is used in many other answers, e.g. see here and here for more cubic examples, and here for $11$th powers.