This assignment is due next Thursday (10/8) but start it over the weekend so we can discuss it some on Tuesday.
x2 + 3y2 = n x2 - 3y2 = n x2 + 5y2 = n x2 - 5y2 = nIn particular, see if you can make a conjecture about which prime values of n allow a solution. Then see how much of that conjecture you can prove (maybe not much)!
This is like the first question on the last assignment. In fact I built it with cut and paste. You now know that your answer should address the various implications among the three statements
x2 + my2 = p has a solution ... has a square root mod p p == ... (mod ...)Prove what you can. Conjecture where you think it appropriate.
Careful! At least one of the conjectures you might be tempted to make has a small counterexample. Look for it.
x2 - 3y2 = +-1 x2 - 5y2 = +-1
You should be able to find a few small solutions by inspection and then find a pattern that allows you to prove that there are infinitely many solutions. It's harder to show that the infinite list you've found includes all the solutions.
x2 - dy2 = +-1has one solution then it has infinitely many.
(In fact it always does have a solution, but we're not ready to prove that yet.)
x2 + y 2 = n (A)and
-1 has a square root (mod n) (B)when n is not prime. (We know the statements are equivalent when n is prime.)
Right after class Professor Wortman suggested thinking about 26. 5 is a square root of -1 (mod 26) and 26 = 52 + 12, so it's not a counterexample to either implication.
|ai,j|
of order 100 with ai,j = i * j.
Prove that if the absolute value of each of the 100! terms
in the expansion of this determinant is divided by 101 then the
remainder in each case is 1.
(This was Problem 1 in the afternoon session for the 17th Putnam
competition, March 2, 1957.)