Symmetric Diophantine Equation | British Mathematical Olympiad 1995 Problem 1

#DiophantineEquations #NumberTheory #MathOlympiad

Here is the solution to BMO2 1995 Problem 1!!

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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.

[BucifalulR]

Calculus is definitely not needed here as that function is clearly strictly decreasing since 1/x is decreasing and hence so is 1+1/x and hence also (1+1/x)^3 since we are only considering x > 0. Also, as others pointed out, the permutations should have been included in the final solution. Still, I liked the solution and presentation, so keep up the good work!

[farhatali]

Your problem selection is very good. Thanks for your effort. Nicely explained.👍

[johnnath]

@Anitabh Kumar. He is quietly spoken, that’s all. It seems that you prefer loud and vociferous. Just listen carefully to every word he says, read with attention every equation he writes and you will (or should) get a sense of the quality of the teacher. I think he is very good.

[hvok]

Wow, this was a really enjoyable problem to work through. I love how many generally problem solving tools were used and how nice of an upper and lower bound there was on a. The solution to this problem from start to finish is pure poetry.

[jkid]

Well, and the permutations of the solutions you gave, since the ordering of a b and c was something you did and not something baked into the problem. That said, lovely video. These are always so immersive.

[pedrojose]

Good morning! I came from Brazil, and I do not have a good english. My apologies.
(a+1)/a * (b+1)/b * (c+1)/c = 2 (i) a, b, c integers greater than 1.
All factors (1+1/a), (1+1/b) and (1+1/c) are grater than 1.
All a, b, c must be greater than 1. IF a =1, e.g., so (1+1/a) =2 and you have xy=1 and x>1 and y>1, no way.
So, a>1, b>1, c>1.
For any solution (a, b, c) we have that all permutations are solutions. So we will find the primitive solutions, where a<=b<=c.
a<=3; because (a+1)^3/a^3 <2 if a >=4.
If a=b=c, it has no integer solution.
If a=b >c we have only a possibility, a=3. a>1 and a <=3. Because if a=2, (a+1)^2/a^2=9/4>2; so (c+1/c)<1 and (c+1/c)>1; no way.
a=b=3 in (i); c=8 and we have a primitive solution and all its pemutations. (3, 3, 8)
Let find solutions where a, b, c are diferents. So a>b>c. As a>1 and a<=3. We have only 2 possibilities: a=2 or a=3.

If a=2 ; 3/2*(b+1)/b * (c+1)/c >= 2; But for b, it is maximum if c=b+1so b<=6. As b>a=2, b=3 or b=4 or b=5 or b=6 are the only possibilities.
a=2, b=3 we have no solutions for c.
a=2, b=4 and (i); c=15
a=2, b=5, and(i); c=9
a=2, b=6, and(i); c=7
So we have more 3 solucions and all their permutations. (2, 4, 15); (2, 5, 9); (2, 6, 7)

If a= 3; 4/3*(b+1)/b * (c+1)/c >= 2; But for b, it is maximum if c=b+1so b<=4. As b>a=3, b=4 is the only possibility.
Solving (i) for a=3 ; b=4, we easily find c=5 ( also a primitive Pythagorean triple and a AP). And we have the last primitive solution.

So (2, 4, 15); (2, 5, 9); (2, 6, 7); (3, 3, 8); (3, 4, 5) and all their permutations, i.e., 27 solutions at all.

[neelg]

Finally a question I can at least follow along without as much prior knowledge or random tricks 😭

[krutarthshah]

Great video! Although instead of calculus I would have gotten the value of a in terms of cbrt(2) which also indeed gives the same bound although some calculation is required

[niceroundtv]

8:50 I'm glad you caught that mistake. This was a problem I was able to solve on my own.

[jamiewalton]

It was a bit of a waste of time checking all of the solutions, since your reasoning above was reversible (e.g., no point where you might have multiplied by a zero factor) and thus already demonstrates the potential solutions you derived are solutions. This is one reason why it's good (and is just generally more rigorous and easier to read) to add implication and iff arrows between statements and equations.

Also (and you kind of admitted this already anyway) but it's pretty trivial to show directly that (1+1/a)^3 is monotone decreasing in a>0 without calculus (since 0<x<y implies 1/x > 1/y thus 1+1/x > 1+1/y and the hence the same after cubing).

To those saying you should note any permutation is also an answer: that's already implicit in stating WLOG a≤b≤c at the start.

Nice problem, thanks for sharing.

[randomizing]

Nice problem, and perhaps can be done also in a following way: multiplying all elements and grouping, we get ab+bc+ca+a+b+c+1=abc => a(bc-(b+c+1))=bc+(b+c+1) => bc-(b+c+1) | bc+ (b+c+1) and also bc-(b+c+1) | bc-(b+c+1) so bc-(b+c+1) also divides the difference => bc-(b+c+1) | 2(b+c+1) => 2(b+c+1)> bc-(b+c+1) => 3(b+c+1)>bc => (b-3)(c-3)<12. Now looking for b∈{2, 3, 4, 5, 6, 7, 8, 9} we obtain the results.. (it is a bit longer but i guess it can be some alternative though i think it is somewhat similar since it's also based on inequalities, even though yours is clearly better involving calculus in number theory problem..) Great channel you have btw!

[nasrullahhusnan]

In 2:08 you wrote 2<=[1+(1/a)]³ w/o any condition on a. How come? For example if a=2, it is true that [1+(1/a)]³=3.375>2 but if a=10 then [1+(1/a)]³=1.331<3

[-basicmaths]

(a+1)/a *(b+1)/b*(c+1)/c=2
Put b=a+1 and c=a+2 then we get
(a+3)/a=2
a+3=2a
a=3, b=4, c=5

[linkinparker]

I don't know how, but I just looked at this problem and the answer immediately clicked in my mind :()

[mcwulf]

I got one set of solutions very quickly by expanding to make (a+1)(b+2)(c+1)=2abc. Then assigning factors so (a+1)=2b, (b+2)=c, (c+1)=a. This gives (3, 4, 5). I presume the others drop out if I break up the RHS differently.

[sachs]

Ptolemy knew the only ways to divide the octave in 3 superparticular ratios are: a fifth, a major third and a diatonic semitone; a fifth, a minor third and a minor tone; a fifth, a septimal subminor third and a septimal super major tone; two fourths and a major tone and a fourth, a minor and a major thirds.

[sandipshrestha]

0:15 what about if a=b=c=1 or 2 or any --> LHS =6 or less

[jackroger]

I think you were wrong when you got
b=1, 2
c=3, 6
because 3×1=3 not 6
Or I just didn't understand?

[prathmeshraut]

U work is great.. Which country u live?

[_wetmath_]

i used simple algebra to convert the equation into ab+ac+bc+a+b+c=abc and then i got stuck and gave up