How can quadratic congruences like $4u^2+10u+128 \equiv 0\pmod{116}$ be solved? I had no problems splitting the congruence into two parts,$\equiv0\pmod4$ and $\equiv 0\pmod{29}$, getting the solutions $u\equiv0, 2\pmod4$ and $u\equiv4,8\pmod{29}$. But I don’t know how to combine these to solve the original congruence mod 116. I suppose the Chinese Remainder Theorem is involved, but I don’t know how to apply it. Any help is greatly appreciated. Thanks!
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2Lots of older questions on this site about implementing the Chinese Remainder Theorem, have a look at some of them. – Gerry Myerson Aug 19 '22 at 07:03
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See the linked dupes for many worked examples. – Bill Dubuque Aug 19 '22 at 07:59
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Chinese Remainder Theorem
Since $\gcd(4,29)=1$, then by Bezout's identity, which can be solved using the extended Euclidean algorithm or by trial and error for small numbers, we get: $$ 4a+29b=1 $$ which implies: $$ \begin{align} A=29b&=1\pmod4\\ B=4a&=1\pmod{29} \end{align} $$ Therefore: $$ N=Ax+By\\ \Updownarrow\\ \begin{align} N&=x\pmod4\\ N&=y\pmod{29} \end{align} $$ So to solve a given pair of congruences modulo $4$ and $29$, you need to do the steps above:
- Determine $a,b$
- Compute $A,B$ (and reduce them modulo 116)
- Pick the pair $x,y$ from your solutions to the two respective congruences
- Compute $N$ (and reduce modulo 116)
Let us do some of it here, then you can finish:
- Trial and error of Bezout: We have $29\cdot 1=1\pmod4$. Therefore $29\cdot3+1=0\pmod4$. And $29\cdot3+1=88=4\cdot22$. So to sum up: $$ \begin{align} 29&=1\pmod{4}\\ 88&=1\pmod{29} \end{align} $$
- Now we pick $x,y$. Let us take $N=2\pmod4$ and $N=8\pmod{29}$, so that we have $x=2$ and $y=8$.
- Finally compute: $$ N=2\cdot29+8\cdot88=66\pmod{116} $$ And if your partial congruences were right, we should have $u=66$ as one solution to the equation. You will get four solutions as a total by combining the possible values of $x,y$ from the two partial congruences.
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