Prove that $(E): x² - 2px + q = 0 $ if and only if $p² - q \in B \subset \mathbb{Z_{11}} $, and determine $B$.

Question

It is clear that: $x² - 2px + q = (x-p)² - p²+q = 0 $

The equation $(E)$ becomes: $ (x - p)² = p² - q $

Also, we have the equation $x²=a$ admits solutions if and only if $a = 0,1, 4, 9$ (Am I correct?)

Which means $(E)$ admits solutions only if $p²-q \in \{0, 1, 4, 9\} $ (Am I correct?)

The next question was to solve the equation $ x^4+3x²+4 = O $ in $\mathbb{Z_{11}}$.

$x^4+3x²+4 = 0 \iff (x²-2)² = -x² - 3 $

I am unable to proceed to solve this because $ x $ is in both sides. How can I proceed?

Thank you.

@Bernard Because $3 \equiv 25 [11] $ and $5 \equiv 16 [11] $? With $25 = 5²$ and $16 = 4² $. — Zouhair El Yaagoubi, Apr 16 '17 at 12:35
Yes. Just compute all squares in $\mathbf Z/11\mathbf Z$ (to speed up the calculations, you can write its elements as $0,\pm1,\dots,\pm5$). — Bernard, Apr 16 '17 at 12:37

Bernard · Accepted Answer · 2017-04-16T13:39:46.733

1

Hint:

Do as for all biquadratic equations: set $y=x^2$, and solve for $\;y^2+3y+4=0$ first. Then check whether the found roots are squares.

*Some details:

The discriminant is $\;\Delta=9-16=-7=4=(\pm2)^2$. On the other hand, $2^{-1}=6$, so $$y=\frac{-3\pm 2}2=\begin{cases}-5\cdot 6=3,\\-1\cdot 6=5.\end{cases}$$

edited Apr 16 '17 at 13:39

answered Apr 16 '17 at 12:13

Bernard

175,478

$y² + 3y + 4 = 0 $ doesn't have roots because $ \Delta = - 7 < 0 $. The set of solutions is the empty set. Correct? – Zouhair El Yaagoubi Apr 16 '17 at 12:45
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Negative numbers are meaningless in a finite field. Furthermore, $-7=4$. – Bernard Apr 16 '17 at 12:49
$ y² + 3y + 4 = 0 \iff (y + 1)^² = y + 3 $. And from the previous question $y +3$ must be one of $0, 1, 3, 4, 5, 9$.
If for example, $ y + 3 = 0 $, we'll have $ y + 1 = 0 $.

I am confused. Could you please explain to me? Thank you.
– Zouhair El Yaagoubi Apr 16 '17 at 12:56
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Why don't you simply use the resolution formulae? $4$ has two square roots, and you only have to compute the inverse of $2$. – Bernard Apr 16 '17 at 13:00
What do you mean by the resolution formulae? – Zouhair El Yaagoubi Apr 16 '17 at 13:04
Didn't you learn formulae for quadratic equations in high-school? – Bernard Apr 16 '17 at 13:05
English is not my primary langague, sorry.
$\Delta = 4 = 2² = (-2)² $, we will have: $y_1 = - \frac{5}{2}$ and $y_{2} = -1$

$ - 1 = 10 $ which cannot be written as $ 10 = x² $.

The equation doesn't have a solution. Is it correct? Thank you for your help.
– Zouhair El Yaagoubi Apr 16 '17 at 13:25
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There's an error: $-1$ is not a root, and $5/2$ has to be simplified. Please see my updated answer. – Bernard Apr 16 '17 at 13:41
So we will have $x² = 3 = 5²$ and $x² = 5 = 4²$, the set of solutions is $ {4,5 }$. Correct? Thank you so very much for your help sir. – Zouhair El Yaagoubi Apr 16 '17 at 13:47
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It's exactly that. – Bernard Apr 16 '17 at 13:50
Thank you for your help sir! – Zouhair El Yaagoubi Apr 16 '17 at 13:51

Prove that $(E): x² - 2px + q = 0 $ if and only if $p² - q \in B \subset \mathbb{Z_{11}} $, and determine $B$.

1 Answers1