Find a solution to $x^2 \equiv 7 \pmod{787}$.
The answer should be between $1$ and $786$ but beyond that I don't know how to do it.
Asked
Active
Viewed 1,317 times
0
-
1Check this http://math.stackexchange.com/questions/1257648/solving-quadratic-equations-in-modular-arithmetic – Jam Oct 16 '15 at 14:41
-
since 787 is a bigger prime, this is hard. But at least you know $x \equiv 7^{788/2} = 7^{394} \mod 787$. – aras Oct 16 '15 at 14:52
-
I looked over it but still don't really understandit – user274933 Oct 16 '15 at 15:02
-
aras why is that useful? – user274933 Oct 16 '15 at 15:02
-
Your prime $787$ is small enough that if you are not doing a test at the time you can make a spreadsheet or small program to quickly find the answer. If you know neither spreadsheets nor programming, you should learn! – Rory Daulton Oct 17 '15 at 22:47
2 Answers
2
"Find a solution to $x^2 \equiv 7 \pmod{787}$"
Here's the solution. We have two solutions, given your constraints, $x=105$ and $x=682$.
$$105^2=11025=3^2 \cdot 5^2 \cdot 7^2 \equiv 7\mod 787$$ $$682^2=2^2 \cdot 11^2 \cdot 31^2 \equiv 7 \mod 787$$
These can be proved to be the only solutions by using proof by exhaustion. If you want a derivation use this answer. In general, this is the Tonelli-Shanks algorithm. Please read up on that and then come back.
We have $p=787$ and $n=7$. We also have $7^{(787-1)/{2}} \equiv 1$. Now we see that,
$$787-1=786=393 \cdot 2=393 \cdot 2^1=Q \cdot 2^S$$
We have $Q=393$ and $S=1$ so we are done. The solution is given by,
$$7^{(787+1)/4} \mod 787=7^{197} \mod 787=105$$
Which can be solved using the Carmichael function and friends. After that,
The other solution is therefore $787-105$. So the solutions are $x=105$ and $x=682$.
Zach466920
- 8,341
-
Well, yes thank you for doing that but why is that the case? Why are those the solutions? – user274933 Oct 16 '15 at 14:59
-
1@user274933 I can show you that they work. That's why they're the solutions. Do you want a derivation? – Zach466920 Oct 16 '15 at 15:06
-
-
I've been trying to understand how to derive it from the last sentence of your answer but it's still tricky. Yes, I would appreciate a derivation. – user274933 Oct 16 '15 at 15:32
-
-
Ok that's fine. I'm also trying to understand why $x = a^{(p+1)/4}$ is a solution to the congruence $x^{2} \equiv a (modp)$. So I think this is related. Anything you could say about that would be appreciated – user274933 Oct 16 '15 at 15:35
-
Given that p is a prime satisfying $p \equiv 3 (mod4)$ and a is a quadratic residue modulo $p$ I mean. – user274933 Oct 16 '15 at 15:36
-
1
Note that I am working in $\mathbb{Z}_{p}$ where $p$ is a prime (i.e. the elements are equivalence classes, so that I don't need to use (mod $p$)).
As easily follows from Fermat's little theorem $a^p= a$ for all $a$ in $\mathbb{Z}_p$. It follows that if $p +1$ is divisible by $4$, and $x^2 = a$, then $( a^{\frac{p+1}{4}})^2 = ((x^2)^{\frac{p+1}{4}})^2= x^{4 \frac{p+1}{4}}=x^{p+1}=x^2=a$. This tells us that if $a$ has square roots they can be computed as $\pm a^{\frac{p+1}{4}}$.
In your example we compute $x=[7]^{197}=[105]$ and the other one as $-x = [682]$.
Nex
- 3,852