From: "John R Ramsden" Newsgroups: sci.math Subject: X^2 + Y , X + Y^2 = Z^2, T^2 solution Date: Fri, 27 Nov 1998 20:58:25 -0800 [Warning: If your newsreader converts spaces to tabs and/or displays in a proportional spacing font or wraps lines then this posting will look like alphabet soup!] A couple of weeks ago someone posted an interesting problem, possibly one of the Russian ones [?]. But due to a hard disk crash I have been off the air until now, hence the late reply. The problem was to prove that there are no natural number (positive integer) solutions of the pair: X^2 + Y, X + Y^2 = Z^2, T^2 You can't solve this just by means of simple inequalities, any more than Fermat's Last Theorem can be proved by these alone (as some misguided souls persist in attempting - see recent postings!) There are an infinite number of rational solutions, including an infinite class of "algebraic" solutions with X + Y = 1/4. There are obviously trivial integer solutions with one of x, y zero and the other the square of an integer, and there is the non-zero integer solution X, Y = -1, -1. But, as will be shown, there are no non-trivial integer solutions. My approach was to start by finding a general rational parametrization and then prove that this does not include any non-trivial integer solutions. Firstly, for any rational pair x, y satisfying x^2 - y^2 = k, for a given rational k != 0, there is a rational value p != 0 for which: x, y = (p^2 + k) / 2.p, (p^2 - k) / 2.p Conversely any rational value of p != 0 yields by these equations a rational solution of the equation. So this is a "general" rational parametrization. (PROOF: If x, y are rational then so also must be x + y, = p say. Since k != 0 we cannot have |x| = |y|, and this implies p != 0. Thus x - y = k / (x + y) = k/p is defined and is also rational. Then solving this pair of linear equations gives x, y as stated.) In the original pair of equations, assuming XY != 0 (X != 0 && Y != 0), we can use this result to conclude that for any rational X, Y, Z, T there must be a rational pair, m, n (both != 0) for which: 2.m.X + Y, X + 2.n.Y = m^2, n^2 Trivial special case: If 2.n = 1 / 2.m then the first and second of these imply resp: 2.m.X + Y = m^2 = 4.m^2.n^2 whence 4.n^2 = 1, and thus 2.m = 2.n = +/- 1. The -ve option implies -X + Y = X - Y = 1/4, which is impossible as the first implies X = Y contradicting X - Y != 0 implied by the second. The +ve option implies X + Y = 1/4. This precludes an integer solution, but allows an infinite number of rational solutions as in this case the original pair reduces to: (4.Y + 1)^2, (2.Y - 1)^2 = 16.Z^2, 4.T^2 Having dealt with these trivial cases, we can assume that 4.m.n - 1 !=0. In this case the preceding pair can be solved for X, Y to obtain a general rational solution as: X, Y = (2.m^2.n - n^2) / (4.m.n - 1), (2.m.n^2 - m^2) / (4.m.n - 1) Z, T = (m^2 + Y) / 2.m, (n^2 + X) / 2.n If in a rational solution X, Y are integers then by the original equations Z, T are also integers. Thus to prove that no non-trivial integer solutions exist we only need to prove that not both of X, Y can be integers. However, the proof is considerably simplified if we also consider Z, T. For any rational pair m, n there must be integers P, Q, R with (P, Q, R) = 1 and R > 0 such that: m, n = P/R, Q/R Substituting these values in the general solution given above yields: Q.(2.P^2 - Q.R) P.(2.Q^2 - P.R) X, Y = ---------------, --------------- R.(4.P.Q - R^2) R.(4.P.Q - R^2) P^2 + R^2.Y Q^2 + R^2.X Z, T = -----------, ----------- 2.P.R 2.Q.R Next let: (Q, R), (R, P), (P, Q) = u, v, w (0 < u, v, w) Since (P, Q, R) = 1, we must have u, v, w coprime (relatively prime in pairs), and thus there must be coprime integers p, q, r for which: P, Q, R = v.w.p, w.u.q, u.v.r (0 < r) and: (p, u) = (q, v) = (r, w) = 1 Replacing these in the above equations gives: w^2.q.(2.v.w.p^2 - u^2.q.r) w^2.p.(2.u.w.q^2 - v^2.p.r) X, Y = ---------------------------, --------------------------- u.v.r.(4.w^2.p.q - u.v.r^2) u.v.r.(4.w^2.p.q - u.v.r^2) w^2.p^2 + u^2.r^2.Y w^2.q^2 + v^2.r^2.X Z, T = -------------------, ------------------- 2.u.w.p.r 2.v.w.q.r If X, Y, Z, T are integers then in view of the equations for Z, T, we see that r must divide w^2.p^2 and w^2.q^2. Since (r, w) = 1 this means that r must divide p^2 and q^2. But since (p, q) = 1 this forces r = 1. Likewise, again assuming X, Y are integers, the equation for X shows that v divides r, and the equation for Y that u divides r. Hence also u = v = 1. Thus the equations for X, Y reduce to: w^2.q.(2.w.p^2 - q) w^2.p.(2.w.q^2 - p) X, Y = -------------------, ------------------- 4.w^2.p.q - 1 4.w^2.p.q - 1 Since (w^2.q, 4.w^2.p.q - 1) = (w^2.p, 4.w^2.p.q - 1) = 1, these equations imply that X, Y are integers if and only if L = 4.w^2.p.q - 1 divides both of M, N = 2.w.p^2 - q, 2.w.q^2 - p. If |p| <= |q| then: |M| <= 2.w.p^2 + |q| (equality iff q < 0) <= 2.w^2.|p|.|q| + 2.|q| - 1 (equality iff w = |p| = |q| = 1) <= 2.w^2.|p|.|q| + 2.w^2.|p|.|q| - 1 (equality iff w = |p| = 1) <= |L| (equality iff 0 < pq) Similarly, if |q| <= |p| then |N| <= |L| with equality in the latter iff (iff = if and only if) the same conditions, with p, q interchanged, hold. Since 0 < T = min (|M|, |N|) <= |L| and |L| divides T, all the conditions, (collectively the same for either case), must apply, i.e.: w, p, q = 1, -1, -1 and this gives the trivial solution X, Y = -1, -1. ============================================================================== From: Kurt Foster Newsgroups: sci.math,rec.puzzles Subject: Re: Russian 74 puzzle Date: 30 Oct 1998 15:47:02 GMT In sci.math John Scholes wrote: . Can both x^2 + y and x + y^2 be squares for x and y natural numbers? No. Since x and y are both positive integers, for x^2 + y to be a square requires y >= 2*x + 1. But similarly, for y^2 + x to be a square, x >= 2*y + 1 is also necessary. These inequalities cannot hold simultaneously if x and y are positive.