30th IMO 1989 shortlist

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Problem 19

a and b are non-square integers. Show that x2 - ay2 - bz2 + abw2 = 0 has a solution in integers not all zero iff x2 - ay2 - bz2 = 0 has a solution in integers not all zero.

 

Solution

Obviously if there is a not-all-zero integral solution to x2 - ay2 - bz2 = 0, then it is also a not-all-zero integral solution to x2 - ay2 - bz2 + abw2 = 0.

Now suppose that x, y, z, w is a not-all-zero integral solution to x2 - ay2 - bz2 + abw2 = 0. Then we have (x2 - ay2) - b(z2 - aw2) = 0 and hence (x2 - ay2)(z2 - aw2) - b(z2 - aw2)2 = 0. But (x2 - ay2)(z2 - aw2) = (xz - ayw)2 - a(yz - xw)2, so X2 - aY2 - bZ2 = 0, where X = xz - ayw, Y = yz - zw, Z = z2 - aw2. It remains to show that X, Y, Z are not all zero.

If Z = 0, then since a is a non-square, we must have w = z = 0. Hence x2 - ay2 = 0, so x = y = 0. Contradiction. So Z must be non-zero.

 


 

30th IMO shortlist 1989

© John Scholes
jscholes@kalva.demon.co.uk
28 Dec 2002
Last corrected/updated 28 Dec 02