0

I want to prove or disprove this statement:

For any $y \in \mathbb{Z}$ and $y \neq \pm 1$, $(y+1)^{2}$ is not divisible by $y$.

The case where $y$ is even can be easily proved. However, I am stuck at the case where $y$ is odd. I try to start with the fact that "$y+1$ is not divisible by $y$" and expect to use something like the Euclid's Lemma. Any idea on how to proceed?

Kwan Yu
  • 3
  • 1
    What's the remainder if you do polynomial division ($(y+1)^2$ divided by $y$)? – Neat Math Jan 06 '21 at 14:50
  • @NeatMath OH YES this explanation is perfect! – Kwan Yu Jan 06 '21 at 14:55
  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Jan 06 '21 at 15:00

5 Answers5

3

Numbers $y$ and $y+1$ are consecutive and thus relatively prime and thus $y\ne \pm1$ can not divide $y+1$ and so it can not divide $(y+1)^2$.

nonuser
  • 90,026
  • Yes they are co-prime, but I still don't get why "$y$ does not divide $y+1$" implies "does not divide $(y+1)^{2}$". If $y$ itself is prime then the explanation is okay but what if $y$ is composite? Does it matter? – Kwan Yu Jan 06 '21 at 15:01
  • If $y$ does divide $y+1$ then each prime which divides $y$ divides also $y+1$. Is that possible? – nonuser Jan 06 '21 at 15:03
  • Oh yes now I see your point! Thanks for the assistance. – Kwan Yu Jan 06 '21 at 15:10
1

I think I'd go back to the definition here. If $y\mid (y+1)^2$ then there is an integer $k$ such that

$$yk = (y+1)^2 = y^2+2y+1$$

so

$$1 = yk - y^2 -2y = y(k-y-2)$$

Therefore $y$ is a divisor of $1$, which is disallowed.

B. Goddard
  • 1
1

If $y$ divides $(y+1)^2=y^2+2y+1$, knowing that $y$ already divides the terms $y^2$ and $2y$, we conclude that $y\mid 1$. Thus, $y=\pm 1$.

0

Just observe that $y$ does not divide any number in the form of $ky+ 1$ since it divides $ky$. Now note that $(y+1)^2 = 1 + ky$ where $k = y + 2$.

Infinity_hunter
  • 5,369
  • 1
  • 10
  • 30
0

Since you tagged it "polynomials" we highlight their key role here.

Note that $\ y\mid y^2\!+\!1\iff y\mid (\overbrace{y^2\!+\!1\bmod y}^{\large \color{#c00}1})\!\iff y\mid \color{#c00}1$

Generally $\ y\mid f(y)\iff y\mid \overbrace{f(y)\bmod y}^{\large \color{#c00}{f(0)}}\iff y\mid \color{#c00}{f(0)}\,$ via Polynomial $\rm\color{#c00}{Remainder}$ Theorem, where $\,f(x)\,$ is any polynomial with integer coefficvients.

Bill Dubuque
  • 272,048