(2017-08-12) Question Six. 29th IMO,
(Canberra, 16 July 1988).
Let a and b be positive integers such that ab+1
divides a2 + b2. Show that
(a2 + b2 ) / (ab+1) is the square of an integer.
This problem was proposed by Stephan Beck (West Germany)
about whom nothing else is known.
The key is to consider the equation for a fixed value of the quotient k
(we know that k is an integer and want to show it's a perfect square).
a2 + b2 = k (ab + 1)
Descent :
From an hypothetical solution (a,b) let's derive a different one (b,c) :
b2 + c2 = k (bc + 1)
Subtracting the above from our original equation, we obtain:
c2 - a2 = k b (c - a)
To have distincts solutions, we rule out c = a and may divide by (c - a):
c + a = k b
Examining the Least Positive Pair of Solutions :
If a is the least possible positive solution for our choice of k,
then a is no greater than b (or else we could swap the names to make it so).
0 < a ≤ b
With c = k b - a ,
the above shows that (c,a) satisfies the same algebraic condition as (a,b).
However, the integer c can't be positive because...
