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arbitrary congruence consider the equation Crelle's Journal cube Determine those Pythagorean Develop different from zero Diophantine analysis Diophantine equation Diophantine system Diophantus divisible domain with respect equation x2+y2 EXERCISES exist Fermat Find a two-parameter form a domain formula fourth number functional equations greatest common divisor Hence hypotenuse impossible in integers infinite descent infinite number integers x integral Pythagorean triangles integral squares integral value last equation lemma means Mirimanoff mod p2 mod q number of solutions numbers whose sum Obtain solutions obvious positive integral solution prime number prime positive integers problem of solving proof prove rational functions rational numbers rational solution rational triangle relation relatively prime integers relatively prime positive respect to multiplication satisfied by integers second member second solution set of numbers Show single solution solution in integers solution of Eq solving Eq square number T+UVD Thence Theory of Numbers tion two-parameter solution variables whence
Page 104 - To find three squares such that the product of any two of them, added to the sum of those two, gives a square.
Page 110 - ... and c. 35. The edges of two hollow cubes differ by 10 centimeters. If a certain quantity of water is poured into the larger cube, there remain 1578 cubic centimeters of space not filled with water. If the second cube contained 142 cubic...
Page 114 - ... has only a finite number of solutions in integers p and q.
Page 4 - They can be got from right-angled triangles2 by dividing the square of one of the sides about the right angle by the square of the other. Let the squares then be The continued product = -J^ffax...
Page 95 - ... which are prime each to each, then one of these integers (say z) and the sum of the other two (say x+y) are both divisible by q. We ,shall give one other theorem which may be demonstrated by elementary means; namely, the following: V. // p is an odd prime and the equation...
Page 113 - ... the product of two sums of eight squares as a sum of eight squares.
Page 72 - University of Illinois. Prove that the difference of the sixth powers of two integers cannot be the square of an integer. SOLUTION BY ELIJAH SWIFT, University of Vermont. In Carmichael's Diophantine Analysis, pp. 70, 71, the impossibility of the equation x' + y' = 2"z