# How to factor $x^3+y^3-91$?

How can I factor $$x^3+y^3-91=0$$ considering that $$x$$ and $$y$$ are both natural numbers? I tried to do it like this $$(x+y)(x^2-x y+y^2)-91=0$$.

• No factorization is possible at all – Mostafa Ayaz Dec 25 '18 at 20:46
• Maybe it's supposed to be : $x^3 - 64 + y^3 - 27 = ...$ – openspace Dec 25 '18 at 20:48
• @MostafaAyaz so you are saying that there are no natural numbers that can prove x^3+y^3=91? – Lyds Dec 25 '18 at 20:49
• Ledja, you factored $x^3 + y^3= (x+y)(x^2-xy+y^2)$ but you didn't factor $91$. $91=7*13$ so you need to rewrite your equation $(x+y)(x^2-xy+y^2)=7*13$. Then you consider few cases: $(x+y)=7$ and $(x^2-xy+y^2)=13$ and so on and see where lead. – user25406 Dec 25 '18 at 21:09
• If your goal is NOT to write $x^3+y^3-91$ as a product of other nonconstant polynomials (which is impossible), then please edit your question so you don't describe what you want falsely as "factor". – Henning Makholm Dec 25 '18 at 21:31

From your comment above I am interpretting a slightly different question than what you posted:

How can we solve $$(x+y)(x^2−xy+y^2)=91$$?

This is observable precisely as you have already done. if $$(x+y)(x^2−xy+y^2)=91$$ and $$x,y$$ are integers then you can find $$x=3$$ and $$y=4$$ there isn't much hunting you have to do in the sense that $$91$$ is a semiprime.

$$x,y$$ natural numbers implies that $$x+y$$ is a positive integer and this means $$x^2-xy+y^2$$ (which is clearly an integer) is also positive.

But now we need to ask "how can we multiply two numbers and find 91?"

Well of course $$91=7\times 13$$. And because $$91$$ is a "semi-prime" we have very little to consider here. You are right in your comment where you suggest we should consider all of the cases $$x+y=1, x+y=7, x+y=13, x+y=91$$.

Let's do it.

Case 1

$$x+y=1 \implies$$ contradiction. We are going to have a contradiction because $$x$$ and $$y$$ are positive integers (This how I am interpreting your usage of the term "natural numbers.")

Case 2

$$x+y=7$$ this should lead you to the answer you've found. $$x=3, y=4$$ and $$x=4, y=3$$ seem like they are the only solutions by just plugging in some values.

Note for the next two arguments that $$x^2-xy+y^2=91 \implies x^2=91-y^2+xy=91-y(x+y)$$.

Case 3

$$x+y=13$$ But then $$x^2-xy+y^2=x^2-13y=91\implies x^2=91-y(x+y)=91-13y=13(7-y)\implies x$$ is a multiple of $$13$$. Contradiction again. $$13|x \implies x$$ is either $$0$$ or $$13$$.

Case 4

$$x+y=91$$ can be defeated in the same type of argument as the previous case.

Here is some graphical interpretation for all this. Now that I've clarified the cases above and how we can argue away the cases I will explain some intuition I have for this problem. I am someone who has spent a little too long considering the curve $$x^m+y^m=c$$ so forewarning: I may have developed a bizarre intuition for this. They call such people Fermat's Last Followers... (no they don't) but here's a graph of what's happening here:

You should be able to identify the curves $$x^3+y^3=91,x+y=1, x+y=7,x+y=13$$. So because of my relationship to the picture I went to investigating $$x+y=7$$ first. . Just to leave some breadcrumbs about this intuition: the black dot is given by $$(91^{1/3}, 91^{1/3})$$ and this is clearly less than $$(5,5)$$ so it's not worth considering any cases where $$x+y>10$$. Maybe this explains my a little too relaxed attitude towards this case $$x+y=13$$ and my almost complete disregard of $$x+y=91$$. This case is somewhat ludicrous after considering what this looks like.

Oh! And one other bit of mathematician lore. Once you find that $$(3,4)$$ or $$(4,3)$$ is the solution we may stop looking for another answer. The smallest number to be written as the sum of cubes in two different ways is $$1729$$ which has entered math nerd lore as a result of a conversation between Hardy and Ramanujan.

• Sorry about the delete/undelete. I handwaved over a few cases that I should have addressed more directly and now I have. – Mason Dec 25 '18 at 21:55
• Plus one for making a constructive interpretation and an effort greater than most would have for a badly stated Question. – hardmath Dec 25 '18 at 21:59
• @hardmath. Well it's Christmas... – Mason Dec 25 '18 at 22:03