A collection of Diophantine problems with solutions

James Matteson9781440060823, 1440060827

1 Diophantine Problem, It is required to find four affirmative integer numbers, such that the sum of every two of them shall be a cube. Solution. If we assume the first^Cx3^)/3-), the second^^x3-y3–z* ), the third=4(-z3+y3+*’), and the fourth=ws-iOM”^-*)5 then> the first added to the second=B8, the first added to the third=)/3, the second added to third=23, and the first added to the fourth=ir Thus four of the six required conditions are satisfied in the notation. It remains, then, to make the second plus the fourth= v3-y3Jrz*=cnbe, say=ic3, and the third plus the fourth^*3- 23=cube, say=?«3. Transposing, we have to resolve the equalities v3–£=w3–if=u?–oi?; and with values of x, y, z, in such ratio, that each two shall be greater than the third. Let us first resolve, in general terms, the equality «’-}-23=w3-|-y3. Taking v=a–b, z=a-b, w-c–d, y=c-d, the equation, after-dividing by 2, becomes a(a2-)-3i2)==e(c2-J-3f72). Now assume a-Sn])–Smq, b=mp-3nq, c=3nr
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