I missed a lecture where they discussed finding solutions to , and was given this problem to try out. Problem is, I don't even know how to start off since I don't even know how to do the above one.
If and a member of Z, show that has either no solutions or exactly 2 solutions mod p.
If it has 2 solutions, show that only one solution, , satisfies .
Okay let me just make sure I'm getting this correctly
we KNOW it has a solution (but we show that if it has a solution, it must has two?) .
so then
So then,
(substituting)
(subtracting from both sides)
(factoring)
So then,
or
which leads to:
?
And then im guessing that holds because if it was greater than half of p, squaring it would mess it up?
And also .
Ah, yes. I forget how important wording is in proofs sometimes.
EDIT:
Just a follow up for anyone else looking for something similar
One of the solutions is between because the field of and since there are exactly 2 solutions or no solutions, then one solutions must cover half the field, and the other cover the other half.
If I am correct in my thinking, barely covered fields on Friday.
Any congruence with a solution has infinitely many solutions in . When a congruence problem says "has two solutions" they mean two solutions that are distinct modulo the number you are working. If you want to be a little more formal then you can say has zero,one, or two solutions in .
Okay, so you want solutions in Now I know.
Let’s analyse the problem this way. Either has no solutions or there is at least one solution in In the first case, we are done; there is nothing to prove. So we’ll suppose there is at least one solution.
Note that if is one solution, then is another solution. Moreover, as , , otherwise we would have which is impossible as is odd. Hence there are at least two solutions in if any solutions exist at all. This would also answer your last query, namely one of them has to be less than or equal to as it is not possible for both and to be greater than
We still need to show that there are no more than 2 solutions. For this, we use a bit of field theory. Consider the polynomial in As is a field, is a unique-factorization domain. In a UFD, any polynomial of degree cannot have more than roots. Hence the quadratic polynomial can have at most 2 roots modulo Hence there are at most 2 solutions to the equation
Combining the results, we see that has exactly 2 solutions whenever it is solvable.