# Download A collection of Diophantine problems with solutions by James Matteson PDF

By James Matteson

1 Diophantine challenge, it's required to discover 4 affirmative integer numbers, such that the sum of each of them will be a dice. resolution. 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 additional to the second=B8, the 1st additional to the third=)/3, the second one additional to third=23, and the 1st further to the fourth=ir hence 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 unravel the equalities v3--£=w3--if=u?--oi?; and with values of x, y, z, in such ratio, that every will probably be more than the 3rd. allow us to first get to the bottom of, regularly 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 suppose a-Sn])--Smq, b=mp-3nq, c=3nr
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26 A. Diophantine equations and integral forms by formula (12) we have Corollary 2. Under a birational G transformation of the form (10), which gives the equivalence of (1) and (9), every class of solutions of equation (9) maps to a class of solutions of equation (1). Now suppose that in equation (1) we have (ϑn , ϑ1 · · · · · ϑn−1 ) = 1 (n > 1). Therefore mn = (ϑn , [ϑ1 , . . , ϑn−1 ]) = 1, and by Theorem 1 equation (1) is equivalent by a birational G transformation to the equation n−1 Ak Xkmk + An Xn = 0, (14) k=1 all of whose rational solutions are given by the formulas: Xk = tk (1 k n − 1), Xn = − mk n−1 k=1 Ak tk An , where tk (1 k n − 1) are rational parameters.

M) be the conjugates of ω, where m is the degree of J . Since J is contained in Q(θ ), we have ω = g(θ ), where g is a polynomial with rational coefficients. Thus G(x) has a zero in common with the polynomial m g(x) − ω(j ) , (13) j =1 which has rational coefficients, and since G(x) is irreducible, it must divide this polynomial. The factors of (13) are relatively prime in pairs, since their differences are non-zero constants. Hence the polynomials H (j ) (x) = G(x), g(x) − ω(j ) are relatively prime in pairs, and since each of them divides G(x), their product must (1 ) In dealing with this case we do not need to exclude the possibility that (ej , n) = 1.

Vn ) for some rational v1 , . . , vn . Thus when K is cyclic, we have three apparently different conditions on f (x) which are in reality equivalent. Corollary to Theorem 2. Let f (x) be a polynomial with integral coefficients, and suppose that every arithmetical progression contains an integer x such that f (x) is a sum of two squares. Then f (x) = u21 (x) + u22 (x) identically, where u1 (x) and u2 (x) are polynomials with integral coefficients. In the particular case of Theorem 2, namely the case K = Q(i), which is needed for this Corollary, our method of proof has much in common with that used by Lubelski [7] in his investigation of the primes p for which f (x) ≡ 0 (mod p) is soluble.