Show that the number of solution to x2 - y2 ≡ a (p) is p-1 if...

Show that the number of solution to x² - y² ≡ a (p) is p-1 if p not divides a and 2p -1 if p divides a, Use u = x+y and v= x-y

Then show that summation y=0 to p-1 ((y^2 +a)/p) = -1 if p not divides a and

= p-1 if p divides a

Expert Solution

Given . Put and . Then (1) becomes

Since we are solving modulo p, x and y can take values from {0, 1, 2, ..., p-1}.

Case 1. divides . (2) implies is divisible by . But divides , and so divides . That is either both. Let . Then . Then either , since all values of x, y lies between 0 and p-1.

The first option gives, implies .

The second option gives values for namely form 1, 2, ...,p-1. Correspondingly, is fixed and defined by .

Thus we have exactly solutions.

Similarly, if . Then . Then either , since all values of x, y lies between 0 and p-1. But the first option is not possible, as all the values of are between 0 and p-1 and there difference can not be equal to . Therefore, we have the second option . This gives values for namely form 1, 2, ...,p-1. Correspondingly, is fixed and defined by . Thus again we have exactly solutions.

Also the case is not possible as the solutions to both do not match.

So in all we have solutions

Colby Messinger answered 1 year ago

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