### 2.6.22 Chinese remainders : `ichinrem, ichrem`

`ichinrem([a,p],[b,q])` or `ichrem([a,p],[b,q])` returns a list `[c,lcm(p,q)]` of 2 integers.
The first number `c` is such that

∀ k ∈ ℤ, d=c+ k × lcm(p,q)

has the properties

d=a (mod p ), d=b (mod q )

If `p` and `q` are coprime, a solution `d` always exists and all the solutions are congruent modulo `p*q`.
Examples :
Solve :

`⎧ ⎨ ⎩` |
`x` | `=` | `3 (``mod` `5)`
`x` | `=` | `9 (``mod` `13)`

Input :

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

or input :

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

Output :

`[-17,65]`

so `x=-17 (mod 65)`
we can also input :

`ichrem(3%5,9%13)`

Output :

`-17%65`

Solve :

`⎧ ⎪ ⎨ ⎪ ⎩` |
`x` | `=` | `3 (``mod` `5)`
`x` | `=` | `4 (``mod` `7)`
`x` | `=` | `1 (``mod` `9)`

First input :

`tmp:=ichinrem([3,5],[4,7])`

or input :

`tmp:=ichrem([3,5],[4,7])`

output :

`[-17,35]`

then input :

`ichinrem([1,9],tmp)`

or input :

`ichrem([1,9],tmp)`

Output :

`[-17,315]`

hence `x=-17 (mod 315)`
Alternative :

`ichinrem([3%5,4%7,1%9])`

Output :

`-17%315`

Remark
`ichrem` (or`ichinrem`)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 :

`ichrem((3x+1)%5,(4x+2)%7,(x+3)%9)`

Output :

`(-17%315``×``x+156%315`

hence `a=-17 (mod 315)` and `b=156 (mod 315)`.