0

let us consider the following theorem enter image description here

i am surprised about this theorem because if we take for instance following equation enter image description here

and if we divide both side by 2, we will get

enter image description here

one solution is 3 and another solution is 7, but there another solution 11, because $2*11=22$, $22-6=16$ and definitely $8$ divides $16$, there is another solution $19=3+16$ because $2*19=38$ $38-6=32$ and again $32$ can be divided by $8$, then why is there exactly $d$ solution?in our case $d=2$

dato datuashvili
  • 9,194

1 Answers1

1

The point is that 11=3 mod 8 as $11=8 \cdot 1 + 3$. In general when you find a solution $x$ in a mod $ n $ you can find infinite solution in the form $ x+tn $ with $ t \in N$

Augusto Matteini
  • 480
  • yes but exactly that is what i am asking, theorem says about finite solution, but there is infinite solution – dato datuashvili Mar 28 '18 at 16:44
  • The point is that they belong to the same modulo class and hence you can consider 11=3. You have finite solution in the sense that they are exactly the same thing. – Augusto Matteini Mar 28 '18 at 16:46
  • could you elaborate a bit more to clarify situation? – dato datuashvili Mar 28 '18 at 16:51
  • Of course, I'll try. Congruence is related to a clear equivalence relation: we say that $ a \equiv b $ $mod$ $n$ if and only if $ n | a-b$ In this enviroment we can define rest class and say that all the numbers that leaves the same rest in the Euclidean division are actually 'equal' to represent a class we only need a member of a class and we usually take the only number smaller of $n$ that belongs to such class. – Augusto Matteini Mar 28 '18 at 16:57
  • that one i know, just i would like to clarify why theorem says that there exactly d solution mod n – dato datuashvili Mar 28 '18 at 16:58
  • i got point now – dato datuashvili Mar 28 '18 at 17:29