Hi this is a hard thing. Solving Diophantine equations with many variables. My mind hang with this. Somebody helps me. This is it:

x+y congruence to 1(mod7)

x+z congruence to 2(mod7)

y+z congruence to 3(mod7)

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- Oct 12th 2009, 02:48 AMmanusformDiophantine equations
Hi this is a hard thing. Solving Diophantine equations with many variables. My mind hang with this. Somebody helps me. This is it:

x+y congruence to 1(mod7)

x+z congruence to 2(mod7)

y+z congruence to 3(mod7) - Oct 12th 2009, 04:48 AMtonio

Write x + y = 1 + 7k , x + z = 2 + 7m , y + z = 3 + 7n , k,m,n integers.

Substract now the 3rd. one from the 1st.:

x - z = -2 + 7(k-n) , and now add this to the second one

2x = 7(k-n+m) ==> x = 0 (mod 7), and from here y = 1 (mod 7), z = 2(mod 7), and the general solution is:

x = 7k

y = 1 + 7m

z = 2 + 7n

k,m,n integers.

Tonio - Oct 12th 2009, 07:04 PMMedia_Man
x+y=1, x+z=2, y+z=3, so x+y+z=(1+2+3)/2=3. Since y+z=3, x=0. It follows that y=1 and z=2. (The same solution holds in modulo for any base.)