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I'm using the polynomial division method and i got a remainder of 32 but how do I explain in words or mathematically how this justifies my solution.

The previous question has answers based on Properties of Divisibility but I want to know how I can use long polynomial division in my method.

Thanks

Noa Even
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Angela
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  • https://math.stackexchange.com/questions/3286664/prove-for-all-integer-n-1-that-if-n-34-then-n5-and-n29-are-copri/3286669#3286669 – nonuser Jul 11 '19 at 12:09
  • From which integral domain ($\mathbb Q[x]$, $\mathbb R[x]$ or $\mathbb C[x]$) the polynomials you have chosen? – MAS Jul 12 '19 at 07:20
  • This is not a dupe of the claimed question (sinc ethe sought divisibility is opposite). – Bill Dubuque Jul 13 '19 at 15:09

7 Answers7

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If $\ b+5\ $ and $\ b^2+7\ $ are not coprime there must be a prime number $p$ with $$b\equiv -5\mod p$$ This gives $$b^2\equiv 25\mod p$$ and because of $$b^2\equiv -7\mod p$$ we have $\ p\mid 32\ $ , which implies $\ p=2\ $. Hence $\ 2\ $ is the only possible common prime factor. But $\ b+5\ $ is odd if $\ b\ $ is a multiple of $\ 32\ $. Hence $\ b+5\ $ and $\ b^2+7\ $ are coprime in this case.

Peter
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$b^2+7 = (b+5)(b-5) + 32$

So any common factor of $b^2+7$ and $b+5$ must also be a factor of $32$.

Since $b$ is a multiple of $32$, both $b+5$ and $b^2+7$ are odd, so common factor is not a multiple of $2$.

Therefore the only common factor of $b+5$ and $b^2+7$ is $1$ i.e. they are coprime.

gandalf61
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By the Euclidean algorithm $\:\overbrace{(b^{\phantom{|}}\!\!\!+\!5,\,\color{#c00}{b^2\!+\!7})\, =\, (b\!+\!5,\,\color{#c00}{32})}^{\!\!\!\!\!\!\!\!\!\large \bmod b+5:\ \ \ b\ \equiv\ -5\ \ \Rightarrow\ \ \color{#c00}{b^{\Large 2}+7\ \equiv\ 32_{\phantom{|}}}}\, =\, \overbrace{(\underbrace{\color{#0a0}{32n^{\phantom{|}}\!\!\!+\!5}}_{\large b \ =\ 32n},\,32)\, =\, (\color{#0a0}5,32)}^{\large \bmod 32:\ \ \color{#0a0}{32n+5\ \equiv\ 5_{\phantom{|}}}} = 1$

Bill Dubuque
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  • The sought "division" is used implicitly in the $!\bmod$ calculations in the Euclidean algorithm (we need only the remainder, not the quotient). You could write these divisions explicitly and use the extended Euclidean algorithm to get a Bezout identity, but the point of using modular arithmetic is that we don't need to do so since we need only the remainder (mod), not the quotients (which play no role here other than to obfuscate the arithmetical essence of the matter). – Bill Dubuque Jul 11 '19 at 15:02
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Say $d= \gcd$, since $d\mid b^2-25$ we have $$d\mid (b^2+7)-(b^2-25)=32$$

so $d\mid b$. But then $d\mid (b+5)-b=5$ and this can only be if $d\mid \gcd(5,32)=1$.

nonuser
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Let $b=32a$ and so $b+5=2^5a+5$ and $b^2+7=2^{10}a^2+7$. The $1^{st}$ term is a linear polynomial in $a$ with coefficient in $\mathbb{R}$ and its root is in $\mathbb{R}$. But the $2^{nd}$ polynomial does not have any real root. Hence it cannot be factorized as a product of linear polynomial with real coefficient. So they are coprime.

MAS
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Let $d$ be a common factor of $b+5$ and $b^2+7$.

Then $d$ divides $(b^2+7)-(b+5)(b-5)=32,$ so if $32|b$ then $d|b$.

But $d|b$ and $d|b+5$ means $d|5,$ and $d|5$ and $d|32$ means $d=1$ since $\gcd(5,32)=1$.

J. W. Tanner
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Let $d = gcd(b+5, b^{2}+7) = gcd(32k + 5, 32^{2}k^{2} + 320k + 32)$, for some integer $k$. Hence, $d | (b^{2} + 7) - b$, which implies $d | b^{2} + 7 - (b + 5)r$, when $r \in Z.$

Now, we want to choose $r$ such that $(b + 5)r = b^{2} + c$, for some integer $c$; specifically, let $r = (b - 5)$ as then $(b + 5)r$ will result in the difference of two perfect squares $b^{2} - 25$ and permit us to express the difference $b^{2} + 7 - (b - 5)r$ as an integer; that is, $d | b^{2} + 7 - (b^{2} - 25) = 32$.

Certainly $d | b^{2} + 7$ as $b^{2} + 7 = 32^{2}k^{2} + 320k + 32$; however, since $b + 5 = 32k + 5$ and $b = 32k$, it follows that $d | b + 5 - b = 5$; and this allows us to conclude that $gcd(5, 32) = 1.$

Therefore, $gcd(b + 5, b^{2} + 7) = 1$; and so, the two integers contained therein are coprime.

DDS
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