 Posted by By G. H. & Wright, E. M. Hardy

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Then (2) and xz= y (p) i mply NO for some integer k. Si nce lxl < JP and I Y I < posit i ve integer, k 1 and = Jp, x2 + y 2 < 2p so k < 2 . Si nce k i s a ( 3) ns desired. This is o ur first enco unter with a member of the c la ss of Diophantine equation s We shall hav e a lot to sa y a bo ut t his e qua tio n for some sma ll va lue s o f l m l nnu arbitrary n. 32 Polynomials We close this section by studying the Diophantine equation x2 _ yz = p. (4) + The seemingly innocuous change of sign makes the problem trivial.

THE GRO UP (f)(n) We are now ready to complete our study of (n). We shall show that (mn) is isomorphic to (m) X (n) when m and n are relatively prime. Then we can find out all about (n) by factoring n as a product of powers of primes and using our knowledge of the structure of the groups (p«) . The route we follow is straightforward and comput ational . 10 to write (n) as a product of cyclic groups. Let n = 2«p�1 • • • p�· be the factorization of n into products of primes. 1 Theorem.

We shall also call this extended homomorphism p. If fe R [x] and r e R, then p(f(r)) p(f)(p(r)). In particular, the function assigning to each polynomial with integral = = = 12. The Algebra of PolyJWmials 27 coefficients the polynomial with the same coefficients reduced modulo n is a homomorphism from Z [x] to Zn [x] . When fe Z [x] we write J for its image in Zn [x] ; J is the polynomial whose coefficients are the coefficients of f modulo n. Iff has an integral root r, then clearly r is a root of J, but the converse is false.