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I've tried a couple times using the euclidean algorithm but can never get it down to 1, I just get a bigger a bigger answer every time I divide without a remainder, I'm not sure if I'm making a mistake somewhere or just solving it incorrectly in the first place

user577215664
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Magitrix Alyxra
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    Note: $4x^2+1=(2x+1)(2x-1)+2$ – J. W. Tanner Sep 08 '20 at 19:15
  • In fact, the ideal $\langle 4x^2 + 1, 2x + 1 \rangle$ of $\mathbb{Z}[x]$ is the unit ideal: first, by other comments/answers, 2 is in the ideal; and then, $(2x + 1) - x \cdot (2) = 1$ is also in the ideal. So, from this, it should be straightforward to find polynomials $a$ and $b$ such that $a(x) (4x^2 + 1) + b(x) (2x + 1) = 1$. – Daniel Schepler Sep 08 '20 at 19:17
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    " but can never get it down to 1" Can you show what you were doing? – fleablood Sep 08 '20 at 20:13

5 Answers5

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Let $d = \gcd(4x^2+1,2x+1)$. Then $d$ divides any linear combination of $4x^2+1$ and $2x+1$. In particular, $d$ divides $r(2x+1)+s(4x^2+1)$ where $r = 1-2x$ and $s=1$.

azif00
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$$\gcd(4x^2+1,2x+1)=\gcd((2x-1)(2x+1)+2,2x+1)=1.$$

Michael Rozenberg
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Use the extended Euclidean algorithm: \begin{array}{r|cr|l} r(x)&u&v&q(x)\\ \hline 4x^2+1&0&1 \\2x+1&1&0&2x-1 \\ \hline 2&-2x+1&1&x \\ 1 &2x^2-x+1 &-x \\ \hline \end{array} whence a Bézout's identity: $$(2x^2-x+1)(2x+1)-x(4x^2+1)=1. $$

Bernard
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    @MichaelRozenberg: Oh!yes. Copy-paste accident. I'll fix it within seconds. Thanks for warning me! – Bernard Sep 08 '20 at 19:35
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" I just get a bigger a bigger answer every time I divide without a remainder"

Then divide with remainder. The goal is to show the final remainder is $1$.

$4x^2 + 1 = P(x)(2x+1) + R(x)$ well when we divide $4x^2 + 1$ by $2x+1$ we get $2x$ with remainder $-2x+1$. That is to say

$4x^2 + 1 = (4x^2 + 2x) - 2x+1= 2x(2x+1) + (-2x + 1)$.

Now we solve $2x+1 = P(x)(-2x+1) + R(x)$ and when we divide $2x+1$ by $-2x+1$ we get $-1$ with remainder $2$. That is to say

$2x+ 1 = (2x-1) + 2 = -(2x+1) + 2$.

And now we sove $-2x +1 = P(x)\cdot 2 + R(x)$ and when we divide $-2x+1$ we get $-x$ with remainder $1$. That is to say

$-2x + 1 = (-2x) + 1 = -x(2) + 1$.

And once we reach $1$ we are done. If we need convincing, we can try to solve $2 = P(x)\cdot 1 + R(x)$ where as $2 = 2\cdot 1 + 0$ we have completed the division and the last remainder, $1$ is the $\gcd$.

So the $\gcd$ is $1$.

Alternatively:

$\gcd(4x^2 + 1, 2x+ 1) = \gcd([4x^2+1]- 2x[2x+1], 2x+1)=$

$\gcd(-2x + 1, 2x+1) =\gcd([-2x+1]+[2x+1], 2x+1) =$

$\gcd(2, 2x+1) = \gcd(2, [2x+1] -x[2])=$

$\gcd(2, 1) = 1$.

fleablood
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$ \underbrace{(\overbrace{(\color{#0a0}{2x})^2\!+1}^{(\large\color{#0a0}{-1})^2\,+\,1},\,2x\!+\!1)}_{\!\!\!\large\color{#0a0}{2x\ \equiv\ -1}\pmod{\!2x+1}}\, =\, \underbrace{(2,\,\color{#c00}2x\!+\!1)}_{\!\!\!\!\!\!\!\large\bmod 2\,:\ \ \color{#c00}{2\ \equiv\ 0}} = (2,1)=1,\, $ by $\ \underbrace{\textstyle (a,b) = (a\bmod b,\,b)}_{\large \text{Euclidean algorithm}}$

Bill Dubuque
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