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As part of solving a problem in elementary number theory, I got to this condition which I'd like to prove.

I went about it as follows:

If $d=\gcd(4k, 2k+1)$ then $4k \equiv 0 \equiv 2k+1\pmod d$ and therefore $0 \equiv 2(2k+1) - 4k \equiv 2\pmod d$, whence $d \mid 2$. Therefore either $d = 1$ or $d = 2$.

However, clearly d cannot be 2 since $2k+1$ is an odd number, therefore it must be that $d = 1$, QED.

This seems to prove the result, however I am bothered by the fact I cannot find a representation of d ($=1$) in the form $1 = ax + by$ for x, y integers where $a=4k$ and $b=2k+1$. I would expect to be able to find such an identity given Bezout's Identity.. what am I missing?

Bill Dubuque
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giorgio
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    Hint: $a$ and $b$ may involve $k$ – Empy2 Apr 16 '23 at 12:16
  • @Empy2 aha I think I follow - this is really finding a gcd of two polynomials (albeit of degree 1), hence the form of Bezout's identity relevant here is one where the co-efficients x and y may be polynonimals themselves (although still only involving whole numbers per my needs) – giorgio Apr 16 '23 at 12:26
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    If you like, yes. But I was thinking the solution for $4$ and $3$ will be different from $8$ and $5$, or $12$ and $7$, and you might spot a pattern in the $a$ and $b$. – Empy2 Apr 16 '23 at 12:33
  • $(\color{#0a0}{1-2k})(1+2k) = 1 - 4k(\color{#c00}k) = $ Bezout equation for $,\gcd(1+2k,4k),,$ by

    $\bmod 2k+1!:\ 2k\equiv -1\Rightarrow (2k^2)\equiv 1\Rightarrow 1/4k\equiv \color{#c00}k.\ $ Or, alternatively

    $!\bmod 4k!:\ \color{#c00}{(2k)^2\equiv 0}\Rightarrow \dfrac{1}{1!+!2k}\equiv\dfrac{1}{1!+!2k}\dfrac{1!-!2k}{1!-!2k}\equiv \dfrac{\color{#0a0}{1!-!2k}}{1!-!\color{#c00}{(2k)^2}}$ by $\rm\color{#c00}{nilpotent}$ simpler multiples

     – Bill Dubuque Apr 16 '23 at 18:06
  • The first method is explained in the first linked dupe, ie. $2k$ is a unit (invertible) hence so too is its (square) power $(2k)^2$ hence so too is its factor $4k$, with inverse being its cofactor $k = (2k)^2/(4k)$. The second method uses a generalization of rationaling the denominator to invert a unit + nilpotent, see the method of simpler multiples. – Bill Dubuque Apr 16 '23 at 18:16
  • Above we used that computing the Bezout equation is equivalent to computing modular inverses, see here the third dupe. – Bill Dubuque Apr 16 '23 at 18:21
  • Or apply the (polynomial) extended Euclidean algorithm as here in the fourth dupe, viewing the gcd args as polynomials in $,k.\ \ $ – Bill Dubuque Apr 16 '23 at 18:32

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Let $x=-(k+1), y=(2k+1)$

$$4kx+(2k+1)y=-4k(k+1)+(2k+1)^2=1$$

MathFail
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