How to solve $ x^2=19 \mod 25$. Modular of prime power.

Question

Is this method correct?

$x^2=19 \mod 25, x^2=44 \mod 25, x^2=69 \mod 25, x^2=94 \mod 25, x^2=119 \mod 25, x^2=144 \mod 25$

Therefore, $x=12 \mod 25, x= 13 \mod 25$.

Or should I have to split $x^2 = 19 \mod 25$ into $x^2= 19 \mod 5, x^2 = 19 \mod 5$. $x^2= 19 \mod 5$ gives $x=2 \mod 5$, $x= 3 \mod 5$. By using $x=2 \mod 5 , x= 3 \mod 5$ I would arrive at the same $x = 12 \mod 25, x= 13 \mod 25$.

Using the second method, I found $x=5k+2$, $x=5k+3$. By substituting this in $x^2=19mod25$ I get $20k+4 = 19 mod 25$ which gives $4k = 3 mod 5$. Therefore, $k = 2$ which provides $x=12$. Similarly, when I substitute $x=5k+3$ in $x^2=19mod25$ I get $x=13$

Are both ways correct?

Please format your question in MathJax. See here for further help. — Infinity_hunter, Jun 11 '21 at 08:33
The first method is certainly correct. I can't tell what the second method does. Certainly $x^2 \equiv 19 \mod 25$ implies that $x^2 \equiv 19 \mod 25$ and that $x \equiv 2, 3 \mod 5$. How does one get from that to the solution in the second method? — Cade Reinberger, Jun 11 '21 at 08:36
We can't comment on the correctness of the 2nd method since you don't say how you lift ("arrive at") the roots mod $5$ to roots mod $25.,$ But it looks a bit fishy based on what you wrote. Please elaborate. — Bill Dubuque, Jun 11 '21 at 08:46
For the 2nd approach, assume that you realize that $x$ must be $\equiv_{5}$ to either $2$ or $3$. Then, you could let $a$ range thru the elements in ${0,1,2,3,4},$ $b$ range thru the elements ${2,3}$ and check $x^2 \pmod{25}$ for $x$ equal to each of the 10 possibilities of $(5a + b)$. — user2661923, Jun 11 '21 at 08:51
@user2661923 Which is the same as using Hensel lifting (but without the generality) — Bill Dubuque, Jun 11 '21 at 08:51
Any comments on the second method please, now that I have elaborated it in the question. Is it right or wrong ? Thank you. — vinoth s, Jun 11 '21 at 09:20
The 2nd method when elaborated is essentially Hensel lifting, which is already discussed here many times, e.g. see the linked dupes. — Bill Dubuque, Jun 11 '21 at 09:55

Martund · Accepted Answer · 2021-06-11T09:34:35.967

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The first method is correct. Essentially, what you're doing is that $$x^2\equiv 144\pmod{25}\\ \implies (x-12)(x+12)\equiv0\pmod{25}$$ Now product of two numbers is divisible by $25$ means either both of them are divisible by $5$ (which is not possible here because their difference is $24$ which is not divisible by $5$) or one of the numbers is divisible by $25$. So the required solution is $x\equiv12\ \text{or}\ 13\pmod{25}$.

Edit:

Now that the question is edited, what OP meant in the second method is clearer now. So that is also right.

edited Jun 11 '21 at 09:34

answered Jun 11 '21 at 08:48

Martund

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It's best to wait for the OP to clarify rather than post answers based on guesses about what the OP intended. Why do you claim that's what the OP is doing? I don't see any hint of that in anything they wrote. – Bill Dubuque Jun 11 '21 at 08:53
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Thank you for the clarification. Using the second method, I found x=5k+2, x=5k+3. By substituting this in $x^2=19 mod25$ I get 20k+4 = 19 mod 25 which gives 4k = 3 mod 5. Therefore, k = 2 which provides x=12. Similarly, when I substitute x=5k+3 in $x^2=19 mod 25$ I get x=13. – vinoth s Jun 11 '21 at 08:58
@vinoths It is important to put that context into the question, not in comments (else your question might get closed for lack of needed context). – Bill Dubuque Jun 11 '21 at 09:00
@BillDubuque Okay I will add it . I am new to this community so getting used to how it works. – vinoth s Jun 11 '21 at 09:04
Any comments on the second method please, now that I have elaborated it in the question. Is it right or wrong ? Thank you. – vinoth s Jun 11 '21 at 09:19
If that's what you meant @vinoths, that is right. – Martund Jun 11 '21 at 09:33
Okay, thanks for the clarification. – vinoth s Jun 11 '21 at 09:35

How to solve $ x^2=19 \mod 25$. Modular of prime power.

1 Answers1