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If we have for example the equation: $x^2+x+2 \equiv 0 \pmod{56}$, I understand that we need to divide to 2 equations:

  1. $x^2+x+2\equiv 0\pmod{8}$
  2. $x^2+x+2 \equiv 0 \pmod{7}$

Then, how we unite the two results for the original equation?

J. W. Tanner
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  • First you have to solve the equations mod 7 and mod 8. Have you done that? – Derivative Jun 30 '21 at 18:57
  • @Derivative Yes, for $mod 8$ is $x\equiv 2, 5$ and for $mod 7$ is $x\equiv 3$ –  Jun 30 '21 at 19:02
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    Since you already mentioned CRT, perhaps take a look at this https://en.wikipedia.org/wiki/Chinese_remainder_theorem#Case_of_two_moduli – Anurag A Jun 30 '21 at 19:02
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    Since $\gcd(8,7)=1,$ if you have a pair $(a,b)$ where $a$ solves (1) and $b$ solves (2), then this pair generates a unique solution mod $56.$ Can find it by just searching all the elements of ${0,1,...,55}$ untill you get one which is $a$ mod $8$ and $b$ mod $7.$ That can be shortened as shown in most intro number theory texts. – coffeemath Jun 30 '21 at 19:07
  • You can use Hensel's Lemma: https://en.wikipedia.org/wiki/Hensel%27s_lemma – Josh Jun 30 '21 at 19:55
  • My first instinct was \begin{align} x^2 + x + 2 & \equiv 0 \pmod{56}\ \implies x^2 + x + 2 & \equiv 56 \pmod{56}\ \implies x^2 + x + 2 & \equiv 112 \pmod{56}\ \implies x^2 + x - 110 & \equiv 0 \pmod{56} \end{align} The last equivalence can be factored over the integers. – N. F. Taussig Jun 30 '21 at 20:12

1 Answers1

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Since you've already mentioned that you have find that $$x\equiv 2,5 \text{ mod } 8 \text{ and } x\equiv 3 \text{ mod } 7 $$ are the solution then you've to find the solution that satisfy both.
Hence for the first congruency if 2 is the solution then so is $2+8n$ hence 10 also satisfy the same.
Now look for the second congruency , 3 is the solution then so is $3+7n$ hence 10 also satisfy the same. Therefore , $$x\equiv 10 \text{ mod } 56$$ Now again go back to first congruency. As 5 is the solution then so is $5+8n$ therefore try plug in the values so that this number is 3 more than a multiple of 7. Plug in $n=5$ gives $45$ which satisfy the first congruency. But also $45=7\cdot 6+3$ hence this is the other solution.
Therefore the solution are : $$x\equiv 10,45\text{ mod } 56$$

Amrit Awasthi
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