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

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 $$a+b=c+d$$

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.