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(i) Find one solution of the congruence $x^{2}-3x+32\equiv 0\pmod {125}$.

(ii) Use your answer to part (i) to find two solutions of the congruence $x^{2}-3x+32\equiv 0\pmod {1000}$.

Answer:

(i) The congruence $x^{2}-3x+32\equiv 0\pmod {5}$ has solutions $r=1$ and $r=2$.

If $r=1$, let $a=1+5q$. Then $f(r)=30, k=6, f'(r)=-1, -q+6\equiv 0\pmod {5}$, so $q=1$ and $a=6$.

Now $6^{2}-3\cdot 6+32\equiv 0\pmod {25}$, so we have a solution mod $25$.

Now take $r=6, a=6+25q$. Then $f(r)=50, k=2, f'(r)=9, 9q+2\equiv 0\pmod {5}$, so $q=2$ and $a=56$.

So $x\equiv 56\pmod {125}$.

OR

If $r=2$, let $a=2+5q$, Then $f(r)=30, k=6, f'(r)=1, q+6\equiv 0\pmod {5}$, so $q=4$ and $a=22$.

Now $22^{2}-3.22+32\equiv 0\pmod {25}$, so we have a solution mod $25$.

Now take $r=22, a=22+25q$. Then $f(r)=450, k=18, f'(r)=41, 41q+18\equiv 0\pmod {5}$, so $q=2$ and $a=72$.

So $x\equiv 72\pmod {125}$.

(ii)

If $x^{2}-3x+32\equiv 0\pmod {1000}$, then $x^{2}-3x+32\equiv 0\pmod {125}$ and $x^{2}-3x+32\equiv 0\pmod {8}$.

Thus, $x\equiv 56$ or $72\pmod {125}$ and $x(x-3)\equiv 0\pmod {8}$, so $x\equiv 0$ or $3\pmod {8}$.

If $x\equiv 56\pmod {125}$ and $x\equiv 0\pmod {8}$, then $x\equiv 56\pmod {1000}$.

If $x\equiv 56\pmod {125}$ and $x\equiv 3\pmod {8}$, then $x\equiv 931\pmod {1000}$.

If $x\equiv 72\pmod {125}$ and $x\equiv 0\pmod {8}$, then $x\equiv 72\pmod {1000}$.

If $x\equiv 72\pmod {125}$ and $x\equiv 3\pmod {8}$, then $x\equiv 947\pmod {1000}$.

(Only two needed)

Above are the answers for these two problems (i) and (ii). But I do not understand how the congruence $x^{2}-3x+32\equiv 0\pmod {5}$ has solutions $r=1$ and $r=2$ in part (i). Also, I do not understand how $k=6$ and why $-q+6\equiv 0\pmod {5}$ in part (i). Lastly, I do not understand how to get $x\equiv 56\pmod {1000}, x\equiv 931\pmod {1000}, x\equiv 72\pmod {1000}, x\equiv 947\pmod {1000}$ in part (ii). Can anyone please explain these?

Bill Dubuque
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Enigma
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    $x^2-3x+32 \equiv x^2-3x+2 =(x-2)(x-1) \pmod 5$. – Robert Shore Mar 30 '23 at 02:44
  • One simple (but inelegant way) : having solved (i), suppose that you identify the value $~r~$ which is an element of $~{0,1,2,\cdots,124}.~$ Then, to attack part (ii), you could simply manually check each of the $~8~$ values $~r + 125k ~: ~k \in {0,1,2,\cdots,7}.$ – user2661923 Mar 30 '23 at 03:45
  • Re previous comment, this does not guarantee that the search will work, at least not without further analysis. That is because there could be more than one solution to part (i). However, the problem's wording suggests that the strategy will work. – user2661923 Mar 30 '23 at 03:48
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    It looks like you were absent from class the day they did Hensel's Theorem and the Chinese Remainder Theorem. Brush up on those two topics, and everything should become clear. – Gerry Myerson Mar 30 '23 at 04:06
  • Robert, I understand how $r=1, 2$ now. But how to get $k=6$ and why $-q+6\equiv 0\pmod {5}$ in part (i)? – Enigma Mar 30 '23 at 20:47
  • Have you had a look at Hensel, Enigma? – Gerry Myerson Mar 31 '23 at 12:38
  • Yes, Gerry. But I still do not really understand how to apply Hensel's theorem and the Chinese Remainder Theorem to this. How does the technique of lifting work in here? – Enigma Mar 31 '23 at 17:24
  • See here and its link for how to Hensel lift a root mod $p=5$ to mod $p^3$ (and search on Hensel's lemma/lifting). See the other dupes for how to use CRT when finding modular roots of polynomials. – Bill Dubuque Apr 09 '23 at 00:25

1 Answers1

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I'll walk you through Hensel.

The general idea is you have a polynomial $f$, a prime $p$, an exponent $n$, and a number $a$ such that $f(a)\equiv0\pmod{p^n}$, and you want a number $b$ such that $f(b)\equiv0\pmod{p^{n+1}}$. You take $b=a+p^nc$, where $c$ is to be determined. We have $$ f(b)=f(a+p^nc)=f(a)+p^ncf'(a)+\cdots $$ where all the terms in the dots involve higher powers of $p$ (this is just expansion in Taylor series), so we want $$ f(a)+p^ncf'(a)\equiv0\pmod{p^{n+1}} $$ Divide through by $p^n$ to get $$ p^{-n}f(a)+cf'(a)\equiv0\pmod p $$ If $f'(a)\not\equiv0\pmod p$, then this congruence has a unique solution $c$.

Now let's apply this to the first situation (although I'm using different letters than you are). We have $f(x)=x^2-3x+32$, $p=5$, $n=1$, $a=1$. We want $b=1+5c$ with $c$ to be determined [my $b$ is your $a$, my $c$ is your q]. We have $f(a)=30$, $p^{-n}f(a)=30/5=6$ [this is $k=6$ in your notation], $f'(x)=2x-3$, $f'(a)=-1$, $6-c\equiv0\pmod5$, $c=1$, $b=6$.

Gerry Myerson
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  • Thank you, Gerry. But I do not understand how to get "If $x\equiv 56\pmod {125}$ and $x\equiv 3\pmod {8}$, then $x\equiv 931\pmod {1000}$. "If $x\equiv 72\pmod {125}$ and $x\equiv 3\pmod {8}$, then $x\equiv 947\pmod {1000}$" in part (ii). Can you please explain how to get these? – Enigma Apr 03 '23 at 00:46
  • That's the other thing I suggested you brush up on, the Chinese Remainder Theorem. That's what you use to combine two (or more) linear congruences into a single linear congruence. So, what happens when you try to apply the Chinese Remainder Theorem here? If you have trouble with it, there must be a dozen questions on this site already with examples done in detail. – Gerry Myerson Apr 03 '23 at 03:34
  • Here is a list of questions tagged chinese-remainder-theorem: https://math.stackexchange.com/questions/tagged/chinese-remainder-theorem – Gerry Myerson Apr 03 '23 at 03:54
  • Making any progress on the Chinese Remainder Theorem? – Gerry Myerson Apr 04 '23 at 12:52
  • Not much, although I attempted to find the answers by applying the Chinese Remainder Theorem. For example, suppose $x\equiv 56\pmod {125}$ and $x\equiv 0\pmod {8}$. Applying the Chinese Remainder Theorem produces: $n=125\cdot 8=1000$. Now we define $N_{k}=\frac{n}{n_{k}}$ for $k=1, 2, ..., r$. Observe that $N_{1}=\frac{1000}{125}=8$ and $N_{2}=\frac{1000}{8}=125$. Then $8x_{1}\equiv 1\pmod {125}$ and $125x_{2}\equiv 1\pmod {8}$. – Enigma Apr 05 '23 at 00:34
  • But I do not know how to find $x_{1}$ and $x_{2}$ from here. – Enigma Apr 05 '23 at 00:36
  • OK, so, the problem is, you don't know how to solve linear congruences $ax\equiv b\pmod m$ (given $a,b,m$, find $x$). That's done by the extended Euclidean algorithm, which again has been the topic of many, many questions here (and appears in every intro Number Theory text ever written, and on dozens of webpages). So, that's one more thing you can look up. – Gerry Myerson Apr 05 '23 at 00:46
  • Consider the linear congruence $8x_{1}\equiv 1\pmod {125}$. Applying the Euclidean Division yields: $125=15\cdot 8+5, 8=1\cdot 5+3, 5=1\cdot 3+2, 3=1\cdot 2+1, 2=2\cdot 1+0$. Then $gcd(8, 125)=1$ and $1\mid 1$. But how should I find $x$ from here? – Enigma Apr 05 '23 at 01:14
  • OK, you've used the Euclidean algorithm, good. Now you need to put the extended Euclidean algorithm into operation. As I said, instructions are all over the place! – Gerry Myerson Apr 05 '23 at 03:05
  • Making any progress, Enigma? – Gerry Myerson Apr 06 '23 at 22:28