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Problem 6 IMO 1988: Let a and b be positive integers
Show that (a2+ b2)/(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 (a2+ b2)/(ab +1) is symmetric in a and b. The problem also has some trivial solutions - for example, a = b = 1 and
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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 (a2+ b2)/(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:
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 = a3 (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:
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.
Dear Unknown
Thank you for posting an alternate way of doing the problem.
It just shows that maths can be good fun.
Regards
Ravi
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