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Is there a way to find the solution to $x \equiv 3 $ (mod $4)$ and $x \equiv 10$ (mod $35$) without trial and error in the scope of elementary number theory? Is there a trivial solution you can immediately see from this? (How would you guess that case?) One solution is 115, but I got this from guessing.

EDIT: To clarify, let me give an example of a problem I did. I solved

$x \equiv 3 $ (mod $4)$, $x \equiv 5 $ (mod $21)$, and $x \equiv 7 $ (mod $25)$ by solving the first two congruences first: $x \equiv 3 $ (mod $4)$, $x \equiv 5 $ (mod $21)$. I guessed that $x = 47$ would work in both by using the first congruence with trial and error. Would there have been an easier way to do this? From this, would there be a trivial answer to the original system provided?

Dominated Convergence Theorem
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2 Answers2

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There are two ways of solving a system of congruences presented in this post. Here is how you can solve your system by solving each equation successively, as in the second answer in the linked post.

Since $x \equiv 10 \pmod{35}$, then $x = 10 + 35a$ for some integer $a$. Substituting this into the first equation yields $$ 3 \equiv x = 10 + 35 a \equiv 2 + 3a \pmod{4} \implies 1 \equiv 3a \implies 3 \equiv a \, . $$ Then $a = 3 + 4b$ for some $b$, so $$ x = 10 + 35a = 10 + 35(3 + 4b) = 115 + 140b $$ and varying $b \in \mathbb{Z}$ produces all possible solutions.

Viktor Vaughn
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You have to perform the extended Euclidean algorithm to have a Bézout's relation between $4$ and $35$: $\;4u+35v=1$. Then $$\begin{cases}x\equiv 3\mod 4\\x\equiv 10\mod 35\end{cases}\iff x\equiv 10\cdot 4u+3\cdot 35v\mod 4\cdot 35$$

Bernard
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