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Science communication is important in today's technologically advanced society. A good part of the adult community is not science savvy and lacks the background to make sense of rapidly changing technology. My blog attempts to help by publishing articles of general interest in an easy to read and understand format without using mathematics. You can contact me at ektalks@yahoo.co.uk

Wednesday 27 May 2020

IMO 1988 Problem 6: General Term Using School-Level Maths


Index of Blogs and Courses
Problem 6 IMO 1988:  Let a and b be positive integers 

Show that (a2b2)/(ab +1) is the square of an integer.
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Problem 6 in the 1988 International Mathematical Olympiad paper has almost reached a legendry status.  The problem is considered extremely difficult to solve - most solutions require a high level of mathematical sophistication or are long and tedious.

Using school level maths, I obtain a general term that may be used to provide a complete list of solutions.  

First we notice that the expression (a2b2)/(ab +1) is symmetric in a and b.  The problem also has some trivial solutions - for example,  a = b = 1 and 
a = 0, b = 1 (b = 0 and a = 1 is also a solution). 

We can also find a special case solution following the simple process described in the slide 


Derivation of the general solution:  In the following, I shall derive a general expression for a and b from which the rest of solutions may be obtained.  This is discussed in the following slides:


As a physicist, I need to check if eq.10 may be simplified further for some special conditions and how that might effect the answer in eq.11.  The obvious case to consider is when x and m are large - this is discussed int he next slide:


It was satisfying for me to be able to provide a general solution for all possible combinations of a and b.  Besides the solutions (a = 0, 1 or 2) and b = a(eq. 3), we have found a new class of solutions such that a must be the cube of a real positive number and b is defined by eq.10 (solutions of the problem in the form asked in IMO 1988). 

2 comments:

Unknown said...

Hi! Great job!
I did something very similar in a different way:
(a²+b²)/(ab+1)=c
a²+b²=abc+c
b²+b(-ca)+(a²-c)=0
Using the quadratic formula:
b=(ac +/- sqr[(ac)²-4(a²-c)])/2
b is a integer, so it is necessary (ac)²-4(a²-c) to be a perfect square:
(ac)²-4(a²-c)=d²
A trivial solution is c=a², so d=ac -> b=(ac +/- ac)/2
b=ac=a³ or b=0
In this case we were chasing only a solution to c, and we saw that c was a perfect square.
A not trivial solution to d is letting a=x^m and c=x^n:
([x^m][x^n])²-4([x^m]²-[x^n])
x^(2m+2n) - 4x^(2m) + 4x^(n)
but (x^(m+n) - 2x^(k))² = x^(2m+2n) - 4x^(k+m+n) + 4x^(2k)
Comparing those terms we see that n=2k and 2m=k+m+n -> m=3k:
a=x^(3k) and c=x^(2k) -> d = x^(5k) - 2x^(k) -> b=[x^(5k) +/- (x^(5k) - 2x^(k))]/2
b=x^(5k) - x^(k)) or b=x^(k)

In summary:
b=a³ or b=0 -> c=a²
a=x^(3k) and b=x^(5k) - x^(k)) or b=x^(k) -> c=x^(2k)
In both cases we get c being a perfect square.

Ravi Singhal said...

Dear Unknown
Thank you for posting an alternate way of doing the problem.
It just shows that maths can be good fun.
Regards
Ravi