An easy way to find solutions of the Diophantine equation A^{3}+B^{3}=C^{3}+D^{3}
aa r X i v : . [ m a t h . HO ] F e b An easy way to find solutions of the Diophantine equation A + B = C + D D. FOSSE, MSc. [email protected]fl.ch
In [1], it is shown how the evenness of the function f ( x ) := ( x + 16 x − + (2 x − x + 42) provides asolution to the diophantine equation A + B = C + D . Let’s generalize this by considering this time g ( x ) :=( a x + a x + a ) + ( b x + b x + b ) . Expanding and simplifying g ( x ) − g ( − x ) gives: g ( x ) − g ( − x ) = 6( a a + b b ) x + 2(6 a a a + 6 b b b + a + b ) x + 6( a a + b b ) x (1)All we have to do is to annihilate the three coefficients of this last polynomial. We want to stay with rationalexpressions , so we start by picking up b from the coefficient of x . This coefficient is zero when b = − a a a + a + b b b and also, the coefficient of x in (1) is zero when b = − a a b . Substitute this expression of b in the previousexpression of b to obtain: b = − a a b and b = (6 a a + a ) b − a a b a (2)Let’s substitute those expressions in the coefficient of x in (1), that is, at this stage, the last one not to benull; we get a a + b b ) = a ( b − a )( a a − a a b − a b )6 a b . Of course, this term is null when a = b or when a = 0 but we also see that a is of degree one on the other factor and that provides a rational expression of it: a a − a a b − a b = ⇒ a = a ( a − b )12 a b . We replace this in the expression of b in (2) and we get finally b = − a ( a − b )12 a b . We can now conclude that h ( x ) := ÅÅ a ( a − b )12 a b ã x + a x + a ã + Å − Å a ( a − b )12 a b ã x − Å a a b ã x + b ã (3)is an even function. Then h ( x ) = h ( − x ) allows us to get rid of the denominators in (3). We also see that a and x behave exactly the same way; so we can drop one of those two variables, say a . Changing x into xy and rearrangingthe terms provides a parametrization of A + B = C + D in terms of binary quadratic forms; i.e.: h ( x ) := (cid:0) q p y + 12 q p xy + q ( q − p ) x (cid:1) + (cid:0) q p y − q p xy − p ( q − p ) x (cid:1) (4)is an even function for the variable x (after having renamed q := a and p := a ). The last step, for the sake ofconcision, is to change y into y qp to get finally: (cid:0) qy + 6 qp xy − q ( p − q ) x (cid:1) + (cid:0) py − pq xy + p ( p − q ) x (cid:1) = (cid:0) qy − qp xy − q ( p − q ) x (cid:1) + (cid:0) py + 6 pq xy + p ( p − q ) x (cid:1) References [1] M.D. Hirschhorn, Ramanujan and Fermat’s last Theorem,
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