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Sample text
If congruence (18) is verified we have just seen that y and z are both divisible by r. Hence from (17) it follows that if either t − x or t + x contains a factor r of the form 4n + 3 it contains that factor to an even power. From this result and the foregoing lemmas it follows at once that integers m, n, p, q exist such that t + x = 2(m2 + q 2 ), t − x = 2(n2 + p2 ), since both t − x and t + x are even. Hence t and x have the form given in (11), while y 2 + z 2 = 4(m2 + q 2 )(n2 + p2 ). Since y and z are even it now follows readily from the corollary to lemma III that for a given t, x, y, z the integers m, n, p, q may be so chosen that y and z are representable in the form given in (11).
Then d is odd. Moreover, the following relations exist, as we see from (9) and (7): θ12 − du21 = 4, θ22 − du22 = 4; (11) u = 21 (θ1 u2 + θ2 u1 ). (12) From Eqs. (11) we see that θ1 and u1 are both odd or both even, and also that θ2 and u2 are both odd or both even. Then from (12) it follows that u is an integer and hence from (10) that t is an integer. We are now in position to prove readily the following theorem: Let D be any positive non-square integer and let σ be any positive integer such that D ≡ 0 mod σ 2 or D ≡ σ 2 mod 4σ 2 .
PROBLEMS INVOLVING A MULTIPLICATIVE DOMAIN 29 numbers of the class form a domain with respect to multiplication. The sets of numbers mentioned also have this further property: from the representation of two numbers in the given form that of their product is readily obtained by means of an algebraic formula. Numbers of the form x2 + y 2 + u2 do not form a domain with respect to multiplication. This may be shown by means of an example. We have 3 = 12 + 12 + 12 , 5 = 22 + 12 + 02 , 21 = 42 + 22 + 12 , while neither 15 nor 63 can be expressed as a sum of three integral squares.