Prove that : $\frac{2k^{7}+1}{3k^{3}+2} \operatorname{reducible}\implies k\equiv 435\pmod{1163}$

Question

Question:

Define $\Omega =\frac{2k^{7}+1}{3k^{3}+2},k\in\mathbb{N}$. Prove that : $$\Omega \operatorname{reducible}\implies k\equiv 435\pmod{1163}$$

My try as follows:

Let $d=\gcd(2k^{7}+1,3k^{3}+2)$

So : $d$ divide $2k^{7}+1$ and $3k^{3}+2$ this mean :

$d$ divide

$$3(2k^{7}+1)-2k^{4}(3k^{3}+2)=3-4k^{4}$$

Then $d$ divide

$4k(3k^{3}+2)+3(3-4k^{2})=8k+9$$

How do I complete it? And where am I atop?

Doubtful. For $k=0,1$, the fraction is irreducible. Didn't you mean $k\not\equiv 435 \pmod {1163}$ ? — GreginGre, Jan 07 '20 at 09:44
I added an answer that uses only very simple mental arithmetic. — Bill Dubuque, Jan 07 '20 at 17:49
In your question 2 months ago I gave for $6$ ways to compute $\dfrac{-9}8\pmod{!1163}\ \ $ — Bill Dubuque, Jan 07 '20 at 18:01
Hmm, I see you also asked the same question 3 months ago. Please don't duplicate your questions. — Bill Dubuque, Jan 07 '20 at 18:03
If you have questions about the prior answers you should ask them in comments there. If you don't get a satisfactory answer then you can ask a new more specific question explaining why the older answers don't work. — Bill Dubuque, Jan 07 '20 at 18:39

score 1 · Accepted Answer · answered Jan 07 '20 at 09:57

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Continuing from your last step, we have, $$d\Big\vert 8(3k^3+2)-3k^2(8k+9)$$ $$\implies d\Big\vert16-27k^2$$ $$\implies d\Big\vert27k(8k+9)+8(16-27k^2)$$ $$\implies d\Big\vert243k+128$$ $$\implies d\Big\vert243(8k+9)-8(243k+128)$$ $$\implies d\Big\vert1163$$ Now, $1163$ is a prime number, and for $\Omega$ to be reducible, $d>1$. Hence, $$d=1163$$ $$\implies 1163\Big\vert 8k+9-3\times1163$$ $$\implies 1163\Big\vert8(k-435)$$ Since $1163$ and $8$ are coprime, we are done.

answered Jan 07 '20 at 09:57

Martund

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very beautiful solution thanks! – Ellen Ellen Jan 07 '20 at 09:59
@EllenEllen, Thank You comments are deprecated by stackexchange (we appreciate the sentiment, though). You can upvote this answer to show your approval and click on the check mark to accept the answer if it solves your problem. – Martund Jan 07 '20 at 10:00
of course , glad you like it – Ellen Ellen Jan 07 '20 at 10:01
Use \mid instead of \Big\vert to get the right spacing – lhf Jan 07 '20 at 13:05

lhf · Answer 2 · 2020-01-07T16:03:11.047

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More systematically:

The extended Euclidean algorithm for gcd gives $$ 1163 = A(2 k^7 + 1)+B(3 k^3 + 2) $$ for some $A,B$ not relevant here.

Let $d=\gcd(2 k^7 + 1,3 k^3 + 2)$. Then $d$ divides $1163$. If fraction is reducible, then $d>1$ and so $d=1163$ because $1163$ is prime.

Now, Euclidean division gives $$ 9(2 k^7 + 1) = (6 k^4 - 4 k)(3 k^3 + 2) + (8 k + 9) $$ Therefore, $d=1163$ divides $8 k + 9$. This is equivalent to $k\equiv 435\bmod{1163}$.

Indeed, to solve $8n+9\equiv 0\bmod{1163}$, find the inverse of $8$ mod $1163$ using the extended Euclidean algorithm. You'll get $727$. Then $n \equiv 727\cdot (-9) \equiv 435 \bmod{1163}$.

edited Jan 07 '20 at 16:03

answered Jan 07 '20 at 11:27

lhf

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$8n+9\equiv 0\pmod{1163}$ then $8n\equiv 1154\pmod{1163}$ how I find value of $n$ ? – Ellen Ellen Jan 07 '20 at 13:19
@EllenEllen, see my edited answer – lhf Jan 07 '20 at 16:03

Bill Dubuque · Answer 3 · 2020-01-07T17:59:43.830

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Hint: $\bmod d = (3k^3\!+\!2,2k^7\!+\!1)\!:\,\ k^3 \equiv -\frac{2}3,\ k^7\equiv -\frac{1}2,\,$ so $\,(-\frac{2}3)^7\equiv k^{21}\equiv (-\frac{1}2)^3,\,$ so $\,0\equiv 3^7-2^{10}\equiv 1163.\,$ Thus $\,k \equiv k^7/(k^3)^3\equiv (-1/2)/(4/9)\equiv -9/8,\, $ and

$\!\!\bmod 1163\!:\,\ 3\left[\dfrac{-3}8\right] \equiv 3\left[\dfrac{1160}8\right]\equiv 3[145]\equiv 435\ \ $ [All easy mental arithmetic]

edited Jan 07 '20 at 17:59

answered Jan 07 '20 at 17:43

Bill Dubuque

272,048

Prove that : $\frac{2k^{7}+1}{3k^{3}+2} \operatorname{reducible}\implies k\equiv 435\pmod{1163}$

3 Answers3