Unsolved Mystery: A Diophantine Equation with an Unusual Set of Integers

[Terry Coates]

Has anyone else spotted an unusual set of three different integers A, B, & C such that
A^n + B^n - C^n = A + B - C > 0 (n > 1 and A x B x C > 0)

I leave the reader to see if they can find this set, or to ask me what they are.

 

[CrazyNinja]

Im assuming here that the 'n' is real and not limited to an integer. Is this allowed or are there restrictions on 'n'?

 

[Terry Coates]

I should have said "an unusual set of four integers" as n is included. A search should find the set quite quickly since none of the integers is as high as 20.

 

[Samy_A]

I should have said "an unusual set of four integers" as n is included. A search should find the set quite quickly since none of the integers is as high as 20.

There are infinitely many solutions for n=2.

 

[Terry Coates]

For n = 2 surely there are no solutions. 3^2 + 4^2 - 5^2 = 0 3 + 4 - 5 = 2

 

[Samy_A]

For n = 2 surely there are no solutions. 3^2 + 4^2 - 5^2 = 0 3 + 4 - 5 = 2

Maybe I misunderstand your question.
Why is, for example, 4²+6²-7²=4+6-7=3 not a solution?

 

[S.G. Janssens]

Maybe I misunderstand your question.
Why is, for example, 4²+6²-7²=4+6-7=3 not a solution?

The product of $A, B$ and $C$ should be positive, I think.

No, sorry, I misread. Your example should be fine.

 

[Terry Coates]

Sorry, you are correct. I should have said n > 2. I think that then there is only one set.

 

[Samy_A]

Sorry, you are correct. I should have said n > 2. I think that then there is only one set.

No, there is more than one set.

 

[Terry Coates]

Many thanks for your speedy replies and research. My set is 16^5 + 13^5 - 17^5 = 12
I'd be pleased to see what others you have discovered.

 

[Samy_A]

$35^3+119^3-120^3=35+119-120=34$

 

[Terry Coates]

Thanks for that.

I actually found my set while searching for the least possible value for A^n + B^n - C^n which in the case of n = 3 the least possible = 2 (with A, B and C relatively prime) I get 64 for n = 4, 12 for n = 5, 69264 n=6, 2697354 n = 7

 

[PeroK]

$35^3+119^3-120^3=35+119-120=34$

That makes it 3-0 to the Belgian.

 

[fresh_42]

That makes it 3-0 to the Belgian.

Or 15-3 in Rugby counts.

 

[Terry Coates]

Seems that these sets can only occur with power 1,2,3 and 5

 

[fresh_42]

Boldings by me.

Seems that these sets can only occur with power 1,2,3 and 5

Sorry, you are correct. I should have said n > 2. I think that then there is only one set.

You already ruled out 1 and 2.

I actually found my set while searching for the least possible value for A^n + B^n - C^n which in the case of n = 3 the least possible = 2 (with A, B and C relatively prime) I get 64 for n = 4, 12 for n = 5, 69264 n=6, 2697354 n = 7

?

 

[Terry Coates]

If I add the condition that A, B and C are to be relatively prime, then my set is probably unique

 

[Samy_A]

If I add the condition that A, B and C are to be relatively prime, then my set is probably unique

Not quite, these also satisfy this additional condition:
3,21,55,56
3,31,56,59
3,49,139,141
3,85,91,111
3,101,291,295

 

[fresh_42]

That makes it 3-0 to the Belgian.

8-0
It's going to be a disaster ...

 

[nasu]

Not quite, these also satisfy this additional condition:
3,21,55,56
3,31,56,59
3,49,139,141
3,85,91,111
3,101,291,295

It does not matter, he will keep adding conditions until his set is unique. :)

 

[WWGD]

8-0
It's going to be a disaster ...

Germany-Brazil in 2014? Is there an isomorphism mapping Samy_A to Germany's soccer team and mapping Terry to Brazil's team?

 

[fresh_42]

Germany-Brazil in 2014? Is there an isomorphism mapping Samy_A to Germany's soccer team and mapping Terry to Brazil's team?

I don't know about Terry but the other comparison ... would you set up an isomorphism between the Bears and the cheese hats?

 

[MostlyHarmless]

It does not matter, he will keep adding conditions until his set is unique. :)

I fail to see how this is inconsistent with Mathematics in general!

 

[fresh_42]

I fail to see how this is inconsistent with Mathematics in general!

It isn't. It only says that either the original problem hasn't been formulated thoroughly enough or carefully enough as, e.g. "How many solutions depending on n does ... have" would have been.

 

[CrazyNinja]

Sorry to disturb your feel people but there are infinitely many solutions for this.

Just take n= any odd integer.
and the condition b= -1; a+c=0.

If it seems lame, forgive me. I think there are infinite solutions for n=even too, but I am looking at it. Will get back!

 

[fresh_42]

Sorry to disturb your feel people but there are infinitely many solutions for this.

Just take n= any odd integer.
and the condition b= -1; a+c=0.

If it seems lame, forgive me. I think there are infinite solutions for n=even too, but I am looking at it. Will get back!

Say $n=2m+1, L(n)=A^n+B^n-C^n, R=A+B-C$.
Then $B=-1$ and $A= -C$ give us $L(n)=2A^n-1$ and $R=2A-1$.
The condition $L(n)=R$ now reads $A^{2m}=1$ which in $ℤ$ is $A,C ∈ \{±1\}$ contradicting the assumption that $A,B,C$ are pairwise different.

 

[Samy_A]

Say $n=2m+1, L(n)=A^n+B^n-C^n, R=A+B-C$.
Then $B=-1$ and $A= -C$ give us $L(n)=2A^n-1$ and $R=2A-1$.
The condition $L(n)=R$ now reads $A^{2m}=1$ which in $ℤ$ is $A,C ∈ \{±1\}$ contradicting the assumption that $A,B,C$ are pairwise different.

Maybe @CrazyNinja meant $A+B=0, C=-1$.
Then, for odd $n$, $A^n+B^n-C^n=A+B-C=1$.

But, probably more interesting, are there solution(s) for $n>5$ with $A, \ B, \ C$ pairwise different positive integers?

No idea. I am not familiar with Diophantine equations, don't know if this has been proven, and if not, how to even start.

 

[fresh_42]

No idea. I am not familiar with Diophantine equations, don't know if this has been proven, and if not, how to even start.

As long as you stay away from book margins everything is alright