Calculus in system of equations

We write the system as $\begin{cases}x^2 + y^2=12^2 + 1^2\\ x^3+y^3=12^3 + 1^3\end{cases}$

This has obvious solutions $x=12,y=1$ and $x=1,y=12$

Now we will show there are no other solutions in positive real numbers. If $(x,y)$ is another solution, without loss of generality we assume that $x>12 >1 >y$ (the case $12>x \geq y >1$ follows by symmetry)

Put $x_1=x^3,y_1=y^3,u_1=12^3,v_1=1^3$

The system becomes

$x^{\frac{2}{3}}_1 - u^{\frac{2}{3}}_1=v^{\frac{2}{3}}_1-y^{\frac{2}{3}}_1$

$x_1 - u_1=v_1 - y_1$

Applying mean value theorem to the function $f(t)=t^{\frac{2}{3}}$ on the intervals $[u_1,x_1]$ and $[v_1,y_1]$ there exists $t_1\in (u_1,x_1)$ and $t_2\in (v_1,y_1)$ such that

$x^{\frac{2}{3}}_1-u^{\frac{2}{3}}_1=\frac{2}{3}t^{\frac{-1}{3}}_1 (x_1-u_1)$

$v^{\frac{2}{3}}_1-y^{\frac{2}{3}}_1=\frac{2}{3}t^{\frac{-1}{3}}_2 (v_1-y_1)$

Consequently $t_1=t_2$ which is impossible since $(u_1,x_1)$ and $(y_1,v_1)$ are disjoint intervals. Hence there are no other solutions.

Final answer $(x,y)=(12,1),(1,12)$ and hence answer is $26$

The number 1729 is famous because it can be written as the sum of two perfect cubes in two different ways: $1729 = 10^3+9^3 = 12^3+1^3.$ Would one of these...? Indeed, $(x,y) = (12,1)$ is a solution!

We must see if there are more solutions. Let $s = x+y$ . Note that $1729 = x^3+y^3 = (x+y)(x^2+y^2-xy) = s(145-xy);$ $145 = x^2+y^2 = (x+y)^2-2xy = s^2-2xy;$ We multiply the first equation by 2 and the second equation by $s$ and subtract: $2\cdot 1729 - 145\cdot s = s(2\cdot145-2xy) - s(s^2 - 2xy) = 2\cdot 145s-s^3.$ Thus we have a cubic equation for the sum $s = x+y$ : $s^3-(3\cdot 145) s+2\cdot 1729=0.$ Because we already have the solution $s = 12 + 1 = 13$ , we factor it out: $(s-13)(s^2+13s-266)=0.$ Equating the second factor to zero gives the additional solutions $s=\frac{-13\pm\sqrt{1233}}{2}.$ The negative sign we discard, because positive values for $x, y$ means that $s > 0$ as well. So focus on the case $x + y = s = \frac{-13+\sqrt{1233}}{2}.$ Using a previous equation, $145 = s^2-2xy = \frac{169+1233-26\sqrt{1233}}{4}-2xy.$ Both $x$ and $y$ are positive precisely if $xy > 0$ . Therefore calculate: $8xy = (169+1233-26\sqrt{1233})-4\cdot 145 = 822-26\sqrt{1233}.$ Because $26^2\cdot 1233 = 833508 > 675684 = 822^2$ , the subtraction will have a negative value.

Thus we have not found any new solutions, and only have $(x,y) = (1, 12)\ \text{and}\ (12, 1)$ . Adding, we find the solution to be 26.