#math practice

[sarahvansloten]

Decimals and Denominators (2024), photographs of found objects (ongoing series)

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[sinesalvatorem]

Tackling Problem 6

So, I was watching an interview with Terence Tao which mentioned his famous Problem Six from the 1988 International Math Olympiad, which he only managed to get one point out of seven for. In the competition, the final problem he had to solve was:

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.

At first I thought this just meant you had to find a single solution at which it is true that:

(a2 + b2) mod (ab + 1) ≅ 0 AND sqrt((a2 + b2) / (ab + 1)) = [some integer]

Which is actually a really easy problem. If you do the obvious thing when testing number-by-number, you get it immediately. If a = 1 and b = 1, then (12 + 12)/(1*1 + 1) = 2/2 = 1, and the root of 1 is of course 1.

But, like, this looked way too easy, so I wanted to know if there were other instances of this forming a perfect square. Then I found that a = 2 and b = 8 does it. Noticing a potential pattern, I checked a = 3 and b = 27. Sure enough, another solution.

At this point, I didn’t want to keep checking these by hand, so I tried the first 25 iterations of b = a3 using a Python script.

a = 1 while(a  <= 25):  b = a**3  if((a**2 + b**2)%(a*b + 1) == 0):    x = (a**2 + b**2)/(a*b + 1)    print("A =", a)    print("B =", b)    print("The result,", x, "is a square of", x**0.5)  a +=1

Upon running it, I found that for all instances of b = a3, the result of the equation is a perfect square. Specifically, the result is a perfect square of a. That was the cool part. (I already suspected this after trying a = 2 and a = 3, but double checking with a computer was trivial enough.)

So then I tried to figure out why that was the case. Firstly, if a3 = b, and a and b are both squared in the numerator, then the numerator is effectively a2 + a6. Next, if the denominator is a*(a3) + 1, then it’s a4 + 1. Finally, the result of the equation is a2. Therefore:

a2 + a6 = (a4 + 1) * a2 #divide both sides by a2# 1 + a4 = a4 + 1

Thus, it is proven that for any positive integers b and a such that b = a3, ab + 1 divides a2 + b2 and a2 + b2 / ab + 1 is the square of an integer.

...Except, that wasn’t what I needed to prove. I needed to prove that for any Z+ a and b such that ab + 1 divides a2 + b2,  a2 + b2 / ab + 1 is a perfect square.

Aaaaand I don’t yet know how to do that. But I’m working on it! So far, I know that not all instances of “ab + 1 divides a2 + b2″ are instances of b = a3, because a = 8 and b = 30 also does it, with a result of 4, which is the square of 2. Meanwhile, a isn’t even a factor of b here: 8 = 2*2*2, while 30 = 2*3*5.

I’m not sure what I’ll need to do to prove the general case, but I’ll work on it some more when I get back from dinner this evening. In the meantime, if anyone has hints/suggestions for how I can tackle this or things I should test, please let me know! Just please don’t write a proof in the notes unless you write it in words and rot13 it.

Edit: More of me working on this problem here.

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[hamlet-the-kat]

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