Show that there do not exist any distinct natural numbers a,b,c,d such that

Question

Show that there do not exist any distinct natural numbers a,b,c,d such that $a^3+b^3=c^3+d^3$ and $a+b=c+d$.

Answer 1

Suppose $a+b=c+d$ and $a^3+b^3=c^3+d^3$. $$a+b=c+d$$ $$(a+b)^3=(c+d)^3$$ $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ $$3ab(a+b)=3cd(c+d)$$ $$ab=cd$$

Let $a+b=c+d=m$ and $ab=cd=n$

a and b are the roots of the quadratic equation $$x^2-mx+n=0$$ by Vieta's relations because a+b=m and ab=n. But c and d are also roots of the equation for similar reasons. But a quadratic equation can have at most two distinct roots.

Hence, a=c or a=d, so a,b,c,d are not distinct.

Answer 2

Mark $a+b=x$ then $b=x-a$ and $d=x-c$. Notice that $x\ne 0$. No we have: $$a^3+(x-a)^3 = c^3+(x-c)^3$$ and thus $$-3x^2a+3xa^2 = -3x^2c+3xc^2$$

so \begin{align}xa-a^2= xc-c^2& \implies x(a-c) = (a-c)(a+c)\\ &\implies x=a+c \\& \implies a+c=a+b \\&\implies c=b\end{align}

A contradiction.