# A collection of Diophantine problems with solutions by James Matteson

By James Matteson

1 Diophantine challenge, it's required to discover 4 affirmative integer numbers, such that the sum of each of them will probably be a dice. answer. If we suppose the first^Cx3^)/3-), the second^^x3-y3--z* ), the third=4(-z3+y3+*'), and the fourth=ws-iOM"^-*)5 then> the 1st further to the second=B8, the 1st extra to the third=)/3, the second one additional to third=23, and the 1st additional to the fourth=ir therefore 4 of the six required stipulations are happy within the notation. It continues to be, then, to make the second one plus the fourth= v3-y3Jrz*=cnbe, say=ic3, and the 3rd plus the fourth^*3- 23=cube, say=?«3. Transposing, we need to get to the bottom of the equalities v3--£=w3--if=u?--oi?; and with values of x, y, z, in such ratio, that every will likely be more than the 3rd. allow us to first get to the bottom of, quite often phrases, the equality «'-}-23=w3-|-y3. Taking v=a--b, z=a-b, w-c--d, y=c-d, the equation, after-dividing via 2, turns into a(a2-)-3i2)==e(c2-J-3f72). Now think a-Sn])--Smq, b=mp-3nq, c=3nr
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Extra info for A collection of Diophantine problems with solutions

Example text

The required set fl, ••. ,fm• Such an element would then form one of Henceforth we shall suppose that l\$s\$g. Let us first assume that the components of a are equal to -s at v o ' 1 at vl' ... • ,vs • As in Chapter I we denote by p the minimal polynomial of the primitive element y in 0 over k[z] and by D the discriminant of P. We shall further assume that each Vi' lSiss, is a finite valuation not in U such that Vi(D) = o. The exceptional cases, when a has a component larger than 1, or when some Vi is either in satisfies Vi (D) > 0, will be dealt with in the next paragraph.

This needs no further comment now, but the construction will be examined in greater detail in Chapter V when we shall require an explicit estimate for the heights of the coefficients of such a basis, as did Coates [10] for the classical case of algebraic numbers. 51 Equations of small genus 2 GENUS ZERO This section is devoted to the construction of an algorithm which enables all the integral solutions of the general equation of genus zero to be determined effectively. Throughout this section we shall denote by F(X,Y) a polynomial with coefficients in K irreducible over the algebraic closure k of K.

A6 in k, not all zero, such that, The conditions v 2 (01) £ -1, v 3 (01) ~ -1 are satisfied, regardless of the values of a l , ... ,a6 , whilst the condition v l (01) £ 2 is equivalent, by consideration of the first Puiseux expansion above of y and t, to the linear equations ° Thus 01 for 03 k. B2 = = a 2 (t+z) = -a 2/t. and B3 Similarly we may determine the possibilities up to factors in k; in fact, if 01 = lit, 02 ~. , ) ) j = 1,2,3, (y-X)/t, is a non-zero element of Furthermore, 1 and 0. The Thue equation 28 Differentiating the second equation we deduce that y' t ~2 + 0, since ~3 0; hence 1 , and X -~lXy/t.