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if $n\in\mathbb{Z}$, then $4$ does not divide $(n^2 - 3)$

I'm not sure how to approach this question, I know how to do questions that involve proving that it does divide but I'm unsure of how to do does not divide. Would I want to use contrapositive as a method of proof?

user158123
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5 Answers5

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$n^2\equiv 0,1 \pmod 4$ so $$n^2-3\equiv 1, 2 \pmod 4$$

Rene Schipperus
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Break it into two cases: even and odd. I'll do the even case (since it's easier but you'll get the idea).

Suppose $n$ is even, then $n = 2m$ for some $m\in\mathbb{Z}$. Then $n^2 = (2m)^2 = 4m^2$. Therefore $n^2 - 3 = 4m^2 - 3$. Can $4$ ever divide this?

Cameron Williams
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  • For the odd case, I get $n^2-3 = 2k-2$ where $k = 2a^2+2a$. Wouldn't 4 be able to divide this in some cases? – user158123 Jun 20 '14 at 02:28
  • Suppose $n = 2m+1$ then $n^2 = (2m+1)^2 = 4m^2+4m+1$. Can you take it from here? (Try cases on $m$ if you need convincing.) – Cameron Williams Jun 20 '14 at 02:32
  • Sorry, this is probably stupid but I don't understand how that implies that 4 can ever divide that. – user158123 Jun 20 '14 at 02:34
  • It's not stupid at all :). So what we get is $n^2 - 3 = 4m^2 + 4m -2$ (following from the last comment). We can rewrite this as $4(m^2+m-1)+2$. This number is a multiple of two that lies between multiples of $4$ - meaning it is not divisible by $4$. – Cameron Williams Jun 20 '14 at 02:36
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    @user158123 Instead of doing an even case and an odd case, try splitting it into four cases: $n = 4m, n=4m+1, n=4m+2, $ or $n=4m+3$. That may make it clearer. – MJD Jun 20 '14 at 02:37
  • @MJD This is a very good suggestion as well. Probably better than mine. – Cameron Williams Jun 20 '14 at 02:37
  • @user158123 Glad we could be of assistance :). Don't hesitate to visit the site to ask us questions if you ever get hung up in the future. – Cameron Williams Jun 20 '14 at 02:45
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Have you done any modular arithmetic? Have a think about what would happen if you worked out the possible values of $n^2 - 3 $ (mod $4$). If 4 is to divide $n^2 - 3 $, then you would need $n^2 - 3 \equiv 0$ (mod $4$). Can this happen?

Mathmo123
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I suppose you could use proof by contrapositive if you really wanted to, but that seems needlessly complicated for something that could so easily be considered with congruences (modular arithmetic), e.g., $47 \equiv -1 \mod 12$ or $47 \equiv 11 \mod 12$.

The simplest modulus would of course be 4 and you're done in the blink of an eye, but to make things a tiny bit more interesting I'm going to use 12; in order for a number to be divisible by 4, it must be 4, 8 or 0 modulo 12. The squares modulo 12 are 1, 4, 9, 4, 1, 0, 1, 4, 9, 4, 1, 0. Then the squares minus 3, taken modulo 12, are 10, 1, 6, 1, 10, 9, 10, 1, 6, 1, 10, 9; none of these are 4, 8 or 0.

Robert Soupe
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Hint $\ $ If $\,f(x)\,$ is a polynomial with integer coef's then, by Polynomial Congruence Rule (PCR)

$\qquad\quad\ \ 4\mid f(n) \iff 4\mid f(\color{#c00}{n\ {\rm mod}\ 4})\ \ $ since $\ \ {\rm mod}\ 4\!:\,\ \color{#c00}{ n'}\equiv n\,\Rightarrow\, f(\color{#c00}{n'})\equiv f(n)\ $ by PCR

Thus if $\ 4\nmid f(n)\ $ for all $\,n \in \{0,1,2,3\}\,$ then $\ 4\nmid f(n)$ for all $\,n\in\Bbb Z.$

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Bill Dubuque
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