For homework I have to produce the proof (algebraic or otherwise) to show that $1729$ HAS to be the smallest taxi cab number. A taxicab number means that it is the sum of two different cubes and can be made with $2$ sets of numbers. I have the list of the next ones and I was wondering if it was linked with the fact that it would have to be $0$ cubed if it got any lower which obviously wouldn't work.
Any help appreciated, thanks in advance!
One can prove that the smallest taxicab number is the smallest product $(6n+1)(12n+1)(18n+1)$ consisting of three primes. This means $n=1$, and $7\cdot 13\cdot 19=1729$. I do not claim that this proof is much better than brute-force.
For positive integers $0 < a < c < d < b < 12$ does the statement $$a^3 + b^3 = c^3 + d^3 = n$$ have any solution(s)? Although a dispositive proof in the negative is referred to, that proof is not presented in detail and it is intimated that the proof may be as much work as the brute force method of answering the question. By "brute force," I presume what is meant is adding essentially all of the various cubes pairwise and looking for matches among the sums. With a little insight, this chore can be reduced to looking at four specific instances, still needing to look at cubes and sums, but so greatly reduced in scope of effort as to be a "gentle force" method. Assuming solutions exist, then:
If $n$ is odd, then $a,b$ have different parities, and $c,d$ have different parities, so $(a+b) \equiv (c+d) mod2$
If $n$ is even, then $a,b$ have the same parity, and $c,d$ have the same parity, so $(a+b) \equiv (c+d) mod2$
Thus, in all cases, $(a+b) \equiv (c+d) mod2$
By Fermat's little theorem, $x^3 \equiv x mod3$ Thus, $(a+b) \equiv (c+d) mod3$
Combining the two results, $(a+b) \equiv (c+d) mod6$
If $(a+b) = (c+d)$, then $n/(a+b) = n/(c+d) = a^2 -ab +b^2 = c^2 -cd +d^2 = (a+b)^2 -3ab = (c+d)^2 -3cd$ Removing the squared terms (which are equal by assumption) and dividing by -3, we get $ab = cd$. If two pairs of integers have the same sum and product, they are identical. So $(a+b) \neq (c+d)$ Hence $$(a+b) \pm 6k = (c+d), k \neq 0$$ This is only possible when $b \ge 10$ These restrictions limit the choices of $(a,c,d,b)$ to (1,2,3,10), (1,8,9,10), (1,2,4,11), (1,8,10,11), (2,3,4,11), and (2,9,10,11). If the both members of either pair, i.e. $(a,b)$ or $(c,d)$, are even, then $8|n$ If both pairs have two even members, then a smaller solution consisting of $(a/2, c/2, d/2, b/2)$ can be found. Within the range of interest, this smaller solution cannot satisfy the $b \ge 10$ requirement. If one pair has two even members then the other pair must have two odd members. If that pair is $(a,b)$ for example, then $a^2 -ab +b^2$, being a combination of three odd numbers, is odd. Hence it must be true that $8|(a+b)$ This restriction eliminates (1,2,4,11) and (1,8,10,11) from consideration. This leaves the four cases (1,2,3,10), (1,8,9,10), (2,3,4,11), and (2,9,10,11) to be examined, and they are easily seen not to satisfy the original statement.
Let $n=a^3+b^3$, and suppose that $\gcd(a,b)=1$. If a prime $p\mid a^3+b^3$, then $$(ab^{-1})^3\equiv_p -1$$ Thus $3\mid \frac{p-1}{2}$, that is, $p\equiv_6 1$.
If we have $n=a^3+b^3=c^3+d^3$, then we can factor $n$ as $$\begin{align}n&=(a+b)(a^2-ab+b^2)\\n&=(c+d)(c^2-cd+d^2)\end{align}$$
Thus we need $n$ to have atleast 3 disctinct prime factors, and so the smallest taxicab number is on the form $$n=(6k+1)(12k+1)(18k+1)$$
