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Prove that for all $n\in\mathbb{N}, (n−1)^3+ n^3 \neq (n+1)^3$

Can someone give me a hint? I can only use properties of natural numbers... don't know how to write up a formal and correct proof for this one...

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yeah... sorry about that. – Alex Lockwood Sep 11 '11 at 18:55
whoa i totally messed up with the edit then :) – user13838 Sep 11 '11 at 18:56
sorry, i forgot to latexify it... :P – Alex Lockwood Sep 11 '11 at 18:57
HINT Assume that the two sides are equal, and get a contradiction. (You may have to group together suitable terms.) – Srivatsan Sep 11 '11 at 19:01
ok, thanks... i'll start with that! – Alex Lockwood Sep 11 '11 at 19:02
up vote 4 down vote accepted

Expand and simplify $(n−1)^3+ n^3 -(n+1)^3$ and prove that it has no zero among the natural numbers.

This can be done as follows: $(n−1)^3+ n^3 -(n+1)^3 = n^3-6 n^2-2 = (n-6) n^2-2$. Now, if this is zero, then $(n-6) n^2=2$. Since the only square that divides $2$ is $1$, we must have $n=1$, which is not a solution of $(n-6) n^2=2$.

Or note that for $n\le 6$, we have $(n-6) n^2\le0$, and for $n\ge7$ we have $(n-6) n^2\ge49$.

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yes... but how do you prove that there are no zeros? using the cubic equation formula seems way too messy... – Alex Lockwood Sep 11 '11 at 19:10
@inradiobows, see my edited answer. – lhf Sep 11 '11 at 19:16
ohhhhhhh... the "since the only square the divides 2 is 1" was key. that was the missing piece of the puzzle :) – Alex Lockwood Sep 11 '11 at 19:20

Suppose the equality hold and just analyze the equation modulo n,

$(n−1)^3+ n^3 = (n+1)^3 \Rightarrow -1 \equiv 1 \pmod{n}$ or $n|2$, but 2 is not a solution. QED

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Note that this is essentially equivalent to applying the Rational Root Test to deduce that any integer root $\rm\:x = n\:$ of an integer coefficient polynomial $\rm\:f(x)\:$ necessarily divides the polynomial's constant term $\rm\:f(0)\:,\:$ i.e. if $\rm\ 0 = f(n)\:$ then mod $\rm\:n:\: \ 0 = f(n)\:\equiv\: f(0)\:,\:$ i.e. $\rm\:n\ |\ f(0)\:.\:$ See my answer. – Bill Dubuque Sep 12 '11 at 3:56

Proof $1.\rm\ \ mod\ 4\:,\:$ the sequence of cubes repeats with period $4\:,\:$ namely $\rm\:\mathbb N^3 =\ 0,\ 1,\ 0,\:-1\:,\: \ldots\:.\:$
Four additions $\rm\:mod\ 4\:$ show that no element in the cycle is the sum of the prior two. $\ $ QED

Proof $2.\ $ $\rm\: (n-1)^3 + n^3 = (n+1)^3\:$ has no solutions $\rm\:mod\ 4\:$ so no integer solutions. This is easy to prove because cubes $\rm\:mod\ 4\:$ have very simple form: for even $\rm\:m = 2\:k\:$ we have $\rm\:m^3 = 8k^3\equiv\: 0\:,\:$ and odd $\rm\:m\:$ are $\equiv \pm1\:,\:$ so $\rm\:m^3\equiv m\:.\ $ Applying this to the equation, considered $\rm\:mod\ 4\:,\:$ we infer

$\begin{eqnarray} &\rm\ \: for&\rm even &\rm n\text{ the equation is}\ &\rm n\!\!-\!\!1\ &+\ \ 0\ &\equiv&\rm\ n\!\!+\!\!1&\rm\: \Rightarrow\ 2\: \equiv\: 0\ \ (mod\ 4)\ \ \ contradiction;\ \ \\ &\rm\ \: for&\rm\ odd &\rm n\ the\ equation\ is &\rm\ \ \ 0\ &+\rm\ \ n\ &\equiv&\rm\ \ \ \ 0&\rm \Rightarrow\ n\: \equiv\: 0\ \ (mod\ 4)\ \ \ contra\ n\ odd. \end{eqnarray}$ QED

NOTE $\ $ The advantage of the second proof is that the derived congruences for cubes are reusable and frequently useful; ditto for $\rm\ odd^2\:\equiv\:\{\pm1,\:\pm3\}^2\:\equiv\: 1\pmod{8}\:.$ Assuming known these ubiquitous congruences, the proof reduces to the above two displayed inferences - which can be derived in less than a minute of trivial mental arithmetic.

This technique of studying integer equations by studying their simpler (finite!) modular reductions in rings of integers modulo $\rm\:m\:$ is quite powerful. It is one of the fundamental problem-solving techniques in number theory and algebra.

Of course one could more simply apply the Rational Root Test to deduce that any integral root $\rm\:x = n\:$ of $\rm\:f(x) = (x+1)^3 - x^3 - (x-1)^3\:$ must divide $\rm\:f(0) = 2\:.\:$ The point of proceeding as above is to illustrate some general techniques which apply to less trivial Diophantine equations.

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Another way to look at it: $(n-1)^3 + n^3$ is approximately $2n^3$, which is too large to be $(n+1)^3$, which is of the same order of magnitude as $n^3$ once $n$ gets to be a certain size.

More rigorously, if $n$ is big enough so that $(n+1)^3 < 2(n-1)^3 = (n-1)^3 + (n-1)^3$, you're also going to have $(n+1)^3 < (n-1)^3 + n^3$. But taking cube roots, $(n+1)^3 < 2(n-1)^3$ is equivalent to $$ (n + 1) < 2^{1 \over 3}(n-1)$$ Solving for $n$ you get $$n > {{2^{1 \over 3} + 1} \over {2^{1 \over 3} - 1}} = 8.6946...$$ So your equality can't hold if $n$ is at least $9$. You can check $n =1,2,...,8$ individually and verify the equality doesn't hold for them either. So it never holds.

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One can do this very quickly with induction for $n \geq 7$. Then there are only a finite number of cases to check.

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This can be proved using various techniques.

One is to compare the left- and right-hand side modulo various small numbers and find one modulus where the two sides are never the same, and then prove it.

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