A solution to a problem of Fermat, on two numbers of which by Euler L.

By Euler L.

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Extra info for A solution to a problem of Fermat, on two numbers of which the sum is a square and the sum of their squares is a biquadrate, inspired by the Illustrious La Grange

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Since 0, some coefficient must be nonzero and (because there are only finitely many coefficients aj) there must be a largest i such that aj i= O. Let n be the largest such i. Hence Proof. f(X) i= f(X) = ao + a1X + a2X2 + ... + anX n where an i= O. We choose g(X) = ~ f(X). an Thus g(X) is monic, has 0' a') a zero and the same degree n as f(X). All the coefficients of g(X) are in F, moreover, since IF is a field. 2 1. If 0' is a zero of 3X3 - 2X + 1, find a monic polynomial with coefficients in Q having 0' as a zero.

Fields of the form IF (0') are essential to our analysis of the lengths of those line segments wliicu can be constructed with straightedge and compass. 1 Famous Impossibilities An Illustration: Q( J2) As Q is a subfield of C, we can consider C as a vector space over Q, taking the elements of C as the vectors and the elements of Q as the scalars. 1 Definition. The set Q(J2) ~ C is defined by putting Q(J2) = {a + bJ2: a,b E Q}. - Thus Q( J2) is the linear span of the set of vectors {I, J2} over Q and is therefore a vector subspace of Cover Q.

3 1. (a) Write down the irreducible polynomial of y'5 over Q and then prove your answer is correct. (b) Write down deg( y'5, Q) . 2. In each case write down a nonzero polynomial f(X) satisfying the stated conditions: (a) f(X) is a monic polynomial over Q with (b) f(X) is a polynomial over Q with momc. J2 J2 as a zero. as a zero but is not (c) f(X) is a polynomial of least degr ee such that f(X) E R[X] and f( V2) = O. (d) f(X) is another polynomial satisfying conditions (c). (e) f(X) E Q[X] and has both J2 and v'3 as zeros .

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