I'm stuck at the ending part of a math exercise on congruences.
I must solve the following system of congruences S:
x≡2 (3)
x≡3 (5)
I was first asked to give the remainders of the division of 3y+2 by 5, with knowing the remainders of the division of y by 5.
Here's what I did:
-If y≡0 (5), then 3y+2≡2 (5)
-If y≡1 (5), then 3y+2≡0 (5)
-If y≡2 (5), then 3y+2≡3 (5)
-If y≡3 (5), then 3y+2≡1 (5)
-If y≡4 (5), then 3y+2≡4 (5)
Here's the part of my exercise I'm stuck with:
I must find the solutions of 3y+2≡3 (5), knowing that each solution x is like x = 15z+8 with z which is an integer. Then, I will need to prove that each integer x like x=15z+8 is a solution of the system S.
I really don't know what to do with this part of my exercise, what shall I do?
Thanks for your answers.
Answer
Hint : 15z+8≡8 ( mod 3 ) and 15z+8≡8 ( mod 5 ) because of 3|15 and 5|15.
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