Show that the Bernoulli number B₁ satisfies B₁ = (-1)" ( 2 + 2d²+1) Σ (mod 1) for any positive integer n where d runs over all of the divisors of n such that 2d+1 is a prime. For example, 1-B₁ = 1/2+3/ 1+ B₂ 1 1/2 + 1/3 + 1 − B₁ = 1/2 + 1/3 + 1/7² 1 1 1+ B4 = 1- B5 = 2 3 2 3 11 1 + B6 = 2 + 3 + 5 +132-B₁ 7 This is known as the Staudt-Clausen theorem, which is found by Staudt (1840) and by Clausen (1840) independently. The latter was published in 'Astronomis- che Nachrichten', the oldest astronomical journal founded in 1821 by H. C. Schumacher (one of the friends and astronomical collaborators of Gauss), as a brief announcement of the result without proof. Later Schwering (1899) gave another proof using 1 1 (n-1)! + + + x + 1 2(x + 1)(x+2) n(x + 1)(x + 2)...(x + n) 1 1 B₁ B₂ = - - + 2x² x3 x³ + = + 17 15 + - lin + + - im B3 x7

1)For each of the following, find an integer value x which satisfies the equation, or explain why there is no solution to the equation. (a)  2x ≡ 3(mod 25) (b)  6x ≡ 4(mod 15)

4. (i) Using the fact that 187 = 11 · 17, show that 2186 = 174 mod 187. This shows that 187 fails the base 2 primality test. (ii) Compute 21911 mod 1911. Use Fermat's little theorem and hand calculations. Does 1911 pass or fail the base 2 primality test? You may use the prime factorization of 1911.