Question

Solve x²+5x+24 ≅ 0 (mod36)

Answer 1

Recall; f(x) ≡ 0 mod mn ,gcd(m,n) = 1 is equivalent to solve the system of congruences f(x) ≡ 0 mod m and f(x) ≡ 0 mod n.

In our case,  As 36 = 4.9 ,  x²+5x+24 ≡ 0 mod 36 is equivalent to x²+5x+24 ≡ 0 mod 4 and x²+5x+24 ≡ 0 mod 9.

Consider x²+5x+24 ≡ 0 mod 4 (or)  x²+x ≡ 0 mod 4 .

Note that x ≡ 0 or x ≡ 3 mod 4 are the residues.

Considering x²+5x+24 ≡ 0 mod 9 (or)  x²+5x+6 ≡ 0 mod 4

Here the residues are x ≡ 6 or x ≡ 7 mod 9.

Now as gcd(4,9)=1 , x ≡ 6, 7 mod 9 gives the possibilities mod 36: {6+9k : k=0,1,2,3} and {7+9h : h=0,1,2,3} which gives {6,15,24,33} and {7,16,25,34} .

Now Checking these against the mod 4 restrictions,

we obtain the solutions as x ≡ 7, 15, 16, 24 mod 36.