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This question already has an answer here:

For all integers $x$ and $y$, if $x^3 + x = y^3 + y$ then $x = y$.

This is what I have done so far:

Proof: Suppose $x$ and $y$ are arbitrary integers. We know that $x^3 + x = y^3 + y$, we want to prove that $x = y$.

So, this is logically making sense, so it is true. any hints please?

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marked as duplicate by Chris Culter, Henning Makholm, Shaun, Zain Patel, user223391 Jul 17 '15 at 18:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jul 17 '15 at 9:01
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Proof:

$x^3+x = y^3+y$

$x^3-y^3+x-y = 0$

$(x-y)(x^2+xy+y^2) + x-y = 0$

$(x-y)(x^2+xy+y^2+1) = 0$

Either$ x-y = 0$ or $(x^2+xy+y^2+1) = 0$

The reason why $(x^2+xy+y^2+1)$ is $((x^2+(\frac{y}{2})^2) + \frac{3}{4}y^2 + 1)$ where for any integer x and y, you have the two squares which can be 0 or greater than 0 + 1. Hence the term cannot be equal to 0. The other responder has made this point which was very obvious to me before.

But the latter is not equal to 0 for integers x and y

Hence $x-y = 0 \Longrightarrow x= y$

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  • $\begingroup$ You can produce $\Rightarrow$ by typing \Rightarrow in math mode; you can produce $\Longrightarrow$ by typing \Longrightarrow in math mode. $\endgroup$ – N. F. Taussig Jul 17 '15 at 9:03
  • $\begingroup$ @N.F.Taussig, Shall use form now on. Learnt something new. Thank you sir $\endgroup$ – Satish Ramanathan Jul 17 '15 at 9:05
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$$ x^3+x=y^3+y\Longrightarrow x^3 - y^3 + x - y = 0\Longrightarrow (x-y)(x^2 + xy + y^2 + 1) =0 $$ But $$ x^2 + xy + y^2 + 1 = \left(x+\frac y2\right)^2 + \frac34 y^2 + 1 > 0, $$ and $x=y$ immediately.

Actually, that follows from that: $f(x)=x^3+x$ is strictly increasing (because $f'(x)=3x^2 + 1 > 0$ for all $x$). Taking derivative is much faster.

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  • $\begingroup$ You don't even need to take derivatives. A sum of two strictly increasing functions is itself strictly increasing, and $x^3$ is hopefully known to be strictly increasing already. $\endgroup$ – Henning Makholm Jul 17 '15 at 9:43
  • $\begingroup$ @HenningMakholm, yes, but it was my first idea $\endgroup$ – Michael Galuza Jul 17 '15 at 10:17
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Note that $f(x) = x^3 + x$ is a strictly increasing function of $x$. As $x$ and $y$ increase, so too does $f(x)$ and $f(y)$.

Assume for contradiction there exists $y\neq x$ such that $f(x) = f(y)$. Then, WLOG, let $x > y$. We have $f(x) = f(y)$, but $f(x) = f(x)$, which means the function $f(\cdot)$ in $[y,x]$ has slope $0$. This is a contradiction to the fact that $f(x)$ is strictly increasing.

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$$x^3-y^3+x-y=(x-y)(x^2+xy+y^2+1)=0$$ has no other solution than $x=y$, because solving the quadratic term for $y$ give a negative discrimiant $x^2-4(x^2+1)$.

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