### 2.6.23 Chinese remainders for lists of integers : `chrem`

`chrem` takes as argument 2 lists of integers of the same size.
`chrem` returns a list of 2 integers.
For example, `chrem([a,b,c],[p,q,r])` returns the list `[x,lcm(p,q,r)]` where `x=a mod p` and `x=b mod q` and `x=c mod r`.
A solution `x` always exists if `p, q, r` are mutually primes, and all the solutions are equal modulo `p*q*r`.
Be carefull with the order of the parameters, indeed :
`chrem([a,b],[p,q])=ichrem([a,p],[b,q])=
ichinrem([a,p],[b,q])`
Examples :
Solve :

`⎧
⎨
⎩`

`x=` | `3 (``mod` `5)`
`x=` | `9 (``mod` `13)`

Input :

`chrem([3,9],[5,13])`

Output :

`[-17,65]`

so, `x=-17 (mod 65)`
Solve :

`⎧
⎪
⎨
⎪
⎩`

`x=` | `3 (``mod` `5)`
`x=` | `4 (``mod` `6)`
`x=` | `1 (``mod` `9)`

Input :

`chrem([3,4,1],[5,6,9])`

Output :

`[28,90]`

so `x=28 (mod 90)`
Remark
`chrem` may be used to find coefficients of polynomial which class are known modulo several integers, for example find ax+b modulo 315=5 × 7 × 9 under the assumptions:

`⎧
⎪
⎨
⎪
⎩`

`a=` | `3 (``mod` `5)`
`a=` | `4 (``mod` `7)`
`a=` | `1 (``mod` `9)`

,

`⎧
⎪
⎨
⎪
⎩`

`b=` | `1 (``mod` `5)`
`b=` | `2 (``mod` `7)`
`b=` | `3 (``mod` `9)`

Input :

`chrem([3x+1,4x+2,x+3],[5,7,9])`

Output :

`[-17x+156),315]`

hence, `a=-17 (mod 315)` et que `b=156 (mod 315)`.