TITLE: Number Theory and Congruency
QUESTION [6 upvotes]: I have the following problem:
$$2x+7=3 \pmod{17}$$
I know HOW to do this problem. It's as follows:
$$2x=3-7\\
x=-2\equiv 15\pmod{17}$$
But I have no idea WHY I'm doing that. I don't really even understand what the problem is asking, I'm just doing what the book says to do. Can someone explain what this problem is asking and what I'm finding? Thanks

REPLY [3 votes]: Starting with the simpler question: I took a number, multiplied it by 
2, added 7, and got 3. What is the number? 
The answer, as you point out, is -2. 
How is the actual question different?: I took an INTEGER x (fractions not allowed), multiplied it by 2, added 7 and got a number y. When I divide y by 17, the remainder is 3. What is x? 
Does the original answer still work? If we take x=-2, divide it by 17 
and take the remainder, we get 15. Now, we multiply by 2 to get 30, 
and add 7 to get y=37. Is this a valid solution? When we divide y by 
17, the remainder is 3. So, yes y=37 is a solution. So the answer is x=-2. 
Are there other solutions? Let's try x=35. If we multiply x by 2 and 
add 7, we get 77. Then, if we divide by 17, and take the remainder, we 
get 9. So, x=35 is not a solution. 
If we try many integers for x, we'll find that x must be one of the following: 
...., -36, -19, -2, 15, 32, 49, .... 
In other words, any multiple of 17 plus 15 will work for x, and ONLY 
those numbers will work. Of course, fractions like 13/2 (ie 6.5) will also work, but we don't include those in modulo problems.
So the "solution set" (not just a single solution) is: 
{x = 15 + 17n, any integer n} 
Another way of saying the above is: "x=15 (mod 17)" [with 3 lines in 
the equal sign, standing for "congruence"]. 
Of course, you don't have to use trial and error to find x. As you 
noted, you actually only need to solve for one value of x, and the 
others will differ by multiples of 17.