Every positive integer is the sum of 4 squares Integers which are not the sum of 4 non-zero squares n =3D 1, 3, 5, 9, 11, 17, 29, 41, 2*4^k, 6*4^k, 14*4^k Theorem - Let M(n) be the number of representations of n as the sum of 4 = squares - M(n) =3D 8 =B7[d|n and d>0 and 4 doesn't divide d] d Sum of 3 squares - It is possible that m and n are each the sum of 3 squares, but mn is = not - Example: 3 =3D 1^2 + 1^2 + 1^2 5 =3D 1^2 + 2^2 + 0^2 15 cannot be the sum of 3 squares Theorem (Legendre/Gauss) A positive integer can be written as a sum of 3 squares if and only if = it is not of the form (8k+7)4^m for some k and m. Waring's Problems - For any k, there is a positive integer s =3D s(k) for some function = s(x) so that every positive integer is the sum of s kth powers - Attempts have been made to generalize s, but have not succeeded Mersenne Primes - The nth Mersenne number is 2^n - 1 - If M[n] is prime, then it is a Mersenne prime Lemma If a^n - 1 is a prime and a > 0, then a has to be 2 and n has to be a = prime Some n values that give Mersenne primes are 2, 3, 5, 7, 13, 17, 19, 31, = 67, 127, 257 (Mersenne generated this list, but was later shown that 2^67 - 1 is = composite) Theorem 3.8 If q is a prime divisor of 2^p - 1, p an odd prime, then q =3D 2kp + 1 = for some k. Proof 2^p congruent to 1 (mod q) ord[q] 2 | p ord[q] =3D p p | q - 1 =3D phi(q) q - 1 =3D mp LHS is even, so m must be even q =3D 2kp + 1 Uncertain if there are infinitely many Mersenne primes Fermat numbers F[n] =3D 2^2^n + 1, F[0] =3D 3, F[1] =3D 5, F[2] =3D 17, F[3] =3D 257, = F[4] =3D 65527 Fermat conjectured that they were all prime, but Euler showed that F[5] = was composite F[5] through F[19] are composite, so conjecture exists that no more = primes exist= --Apple-Mail-1--738558314 Content-Transfer-Encoding: quoted-printable Content-Type: text/plain; x-unix-mode=0644; name="[MA490]7-27-04.txt" Content-Disposition: attachment; filename="[MA490]7-27-04.txt" Let x be in the set of positive reals and =B9(x) by the number of primes = p with 0 < p < x. The Gauss Conjecture states that lim[n->=B0] =B9(n)/(n = log n) =3D 1. It was proved in 1896 by Hadamard and de Valle Poussin. Dirichlet's Theorem If (a,b) =3D 1, a and b > 0, then the sequence {an+b}[n=3D1;=B0] = contains infinitely many primes. Goldbach's Conjecture Every even integer n>2 is the sum of 2 primes. = --------------------------------------------------------------------------= ---- nth triangular number is n(n+1)/2 Square triangular numbers are squares that can be represented = triangularly 8n^2 =3D (2m+1)^2 - 1 for some m and n (mth triangle, nth square) x =3D 2m+1 y =3D 2n x^2 - 2y^2 =3D 1 As it turns out, equation solutions for each cause one for the other (3+2=C32)^n (3-2=C32)^n =3D (x[n]+y[n]=C32)(x[n]-y[n]=C32) =3D 1 x y m n n^2 -- -- -- -- --- 3 2 1 1 1 17 12 8 6 36 99 70 49 35 1225 577 408 288 204 41616 Theorem Every solution to x^2 - 2y^2 =3D 1 is obtained by letting x[n] + = y[n]=C32 be the same as (3 + 2=C32)^n and taking x =3D x[n], y =3D y[n]. Every square triangular number n^2 =3D m(m+1)/2 is given by m =3D = (x[n]-1)/2 and n =3D y[n]/2, with x[n] and y[n] as in the first part. Proof Enough to check that every solution (u,v) to x^2 - 2y^2 =3D 1 is of = the form where u =3D x[n], v =3D y[n] for some n. u + v=C32 =3D (3 + = 2=C32)^n for some n. If u is 3, then v must be 2. Claim: u + v=C32 =3D = (3 + 2=C32)(s + t=C32) with s^2 - 2t^2 =3D 1, 0 < s < u and t > 0. If s = =3D 3 and t =3D 2, we're done. If not, repeat the process. u =3D 3s + = 4t, v =3D 2s + 3t from the claimed equation, resulting in s =3D 3u - 4v = and t =3D -2u + 3v. s^2 - 2t^2 =3D 1 is proven by plugging in these = values of s and t. u^2 =3D 1 + 2v^2 > 2v^2, so u > v=C32, so s =3D = 3u-4v > (3=C32-4)v > 0. u^2 > 9, so we have 9u^2 =3D u^2 + 8u^2 > 9 + = 8u^2, so 9u^2 - 9 > 8u^2, so v > 2u/3, so t =3D -2u + 3v > -2u + 3(2u/3) = =3D 0, so t > 0 and u > s. Pell's Equation (also known as Bramahgupta-Bhaskara equation) x^2 - Dy^2 =3D 1, with D not a perfect square.= --Apple-Mail-1--738558314 Content-Transfer-Encoding: quoted-printable Content-Type: text/plain; x-unix-mode=0644; name="[MA490]7-29-04.txt" Content-Disposition: attachment; filename="[MA490]7-29-04.txt" Pell's Equation: D>0, not a perfect square. x^2 - Dy^2 =3D1 Theorem Pell's Equation always has a positive integer solution. Furthermore, = if {x,y} is the solution with smallest positive first coordinate, then = every solution is given by (x[n],y[n]) with (x[1]+y[1]=C3D)^n =3D = x[n]+y[n]=C3D Example x^2 - 13y^2 =3D 1 11^2 - 13(3^2) =3D 4 and 119^2 - 13(33^2) =3D 4 (11+3=C313)(11-3=C313) =3D 4, (119+33=C313)(119-33=C313) =3D 4 (119-33=C313)/(11-3=C313) =3D (22-6=C313)/4 =3D 11/2 - (3/2)=C313 A solution has been found, but not an integer solution Continuing the process on another set, we try 14159^2 - 13(3927^2) =3D = 4 Dividing through gives 649 - 180=C313, which is the smallest Dirichlet's Diophantine Approximation Theorem Suppose D>0 and is not a square. There are infinitely many pairs = {x,y} with |x - y=C3D| < 1/y. |x^2 - D y^2| < 3 =C3D when |x - y =C3D| < 1/y x^2 - Dy^2 =3D M Since there are infinitely many, we can choose {X[1],Y[1]} and = {X[2],Y[2]} such that X[1] congruent to X[2] (mod M), Y[1] congruent to = Y[2] (mod M). Let x + y=C3D =3D (x[1]-Y[1]=C3D)/(X[2]-Y[2]=C3D). This = results in the following equation: (X[1]X[2]-DY[1]Y[2])/M + = ((X[1]Y[2]-X[2]Y[1])/M)=C3D.= --Apple-Mail-1--738558314 Content-Transfer-Encoding: quoted-printable Content-Type: text/plain; x-unix-mode=0644; name="[MA490]7-30-04.txt" Content-Disposition: attachment; filename="[MA490]7-30-04.txt" #13 --- Show that 2^2^n + 7 is composite for infinitely many n. phi(11) =3D 10 =3D 2*5, and 2^2 congruent to 4, 2^5 congruent to -1 = (both mod 11). 2 is a primitive root of 11. Therefore, 2^2^n congruent to 2^2 (mod 11) means that 2^n congruent to 2 = (mod 10) or 2^(n-1) congruent to 1 (mod 5), so n-1 congruent to 4 (mod = 5) or n congruent to 0 (mod 5). Together, then, 2^2^n + 7 must be composite. #14 --- Want to show that 2^2^n + 1 congruent to either -1 or -4 (mod 9). This = means that 2^2^n congruent to 7 or 4 (mod 9). 2 is a primitive root of 9. 2^2^n congruent to 2^x (mod 9) iff 2^n congruent to x (mod 6). 2^n = congruent to 1, 2, 4, 2, 4, 2.... for increasing powers of n, so it must = be congruent to 2 or 4. Therefore 2^2^n must be congruent to 2^4 or 2^2 = (mod 9), which is 7 or 4 (mod 9). #36 --- Suppose that n is odd and has unique representation n =3D a^2-b^2 with a = and b>0. Show that n is a prime or n =3D p^2 for some prime p. n =3D (a+b)(a-b), a+b > a-b. a+b =3D m a-b =3D p ------- a =3D (m+p)/2 b =3D (m-p)/2 = --------------------------------------------------------------------------= ---- Theorem 3.2 (Dirichlet) p is prime, 1 =B2=CAa =B2=CAp-1. x^2 congruent to a (mod p) is = unsolvable if (p-1)! congruent to a^((p-1)/2) (mod p) and solvable if = (p-1)! congruent to -(a^((p-1)/2)) (mod p). = --------------------------------------------------------------------------= ---- Theorem: -------- a.) p is an odd prime, (-1\p) =3D (-1)^((p-1)/2) b.) (2\p) =3D (-1)^((p^2-1)/8=