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 $a+b=c+d$
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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.