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 :

$$\tt\left \{ \begin{array}{rl} x=&3\ (\bmod\ 5)\\ x=&9\ (\bmod\ 13) \end{array}\right.$$

Input :

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

Output :

[-17,65]

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

$$\tt\left \{ \begin{array}{rl} x=&3\ (\bmod\ 5)\\ x=&4\ (\bmod\ 6) \\ x=&1\ (\bmod\ 9)\end{array}\right.$$

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:

$$\tt\left \{ \begin{array}{rl} a=&3\ (\bmod\ 5)\\ a=&4\ (\bmod\ 7) \\ a=&1\ (\bmod\ 9) \end{array}\right.$$,    $$\tt\left \{ \begin{array}{rl} b=&1\ (\bmod\ 5)\\ b=&2\ (\bmod\ 7) \\ b=&3\ (\bmod\ 9) \end{array}\right.$$

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).