Prove a^2 + b^2 = 3(s^2 + t^2) implies both a and b must are divisible by 3

[Ceres629]

I'm stuck on a seemingly simple part of a proof (a proof showing there are no non zero solutions of the equation a^2 + b^2 = 3(s^2 + t^2)

at one step it says if

a^2 + b^2 = 3(s^2 + t^2) this implies

both a and b must be divisible by 3.
I tried to prove this myself but have had no luck...

any ideas?

 

[CRGreathouse]

If a was divisible by 3 then a^2 would be divisible by 3 (and 9). If b wasn't, then b^2 wouldn't be and so of course a^2 + b^2 couldn't be. Now all you need to show is that if neither of the two are divisible by 3 then their sum can't be. Consider the equation mod 3 (or 9) and you should find the answer.

 

[Ceres629]

I got as far as needing to show that if both a and b are not divisble by three then a^2 + b^2 must not be divisible by 3, but that's exactly where i got stuck,all attempts to show this have proved dead ends :(

*edit*

what i have managed to show is that if:

a not divisible by 3 => a%3 = 1 or 2

if a%3 = 1
a^2 % 3 = 1

if a%3 = 2
a^2 % 3 = 1

thus (a^2 + b^2) % 3 will always be 2 if a%3 and b%3 are non zero

this seems sufficient...

 

[Gokul43201]

Yes, that is sufficient.

This leaves you with the final option: a and b are both divisible by 3.

 

[Ceres629]

Ah okay, thanks for the pointer GreatHouse and thanks for the confirmation Gokul.

 

[al-mahed]

a^2 == -b^2 mod 3

if gcd(a,3)=1 ==> a^2 == 1 == -b^2 mod 3 ==> gcd(b,3)=1

but b^2 == -1 mod 3 is a contradiction ==> a and b must be divisible by 3

 

[robert Ihnot]

Ceres: a^2 + b^2 = 3(s^2 + t^2) this implies
both a and b must be divisible by 3.

Then if a and b are both divisible by 3, we now have 9(a'^2+b'^2) = 3(s^2+t^2), and this leads us to: 3(a'^2+b'^2) =s^2+t^2.

This looks like the original problem reversed!(?)

 

[mathwonk]

just work mod 3. i.e. the equation says you have a^2 +b^2 = 0 mod 3. trying a,b = 0,1,2, one sees nothing works except a=b=0.

so in fact the c^2+d^2 part is a red herring.

 

[Gib Z]

This looks like the original problem reversed!(?)

I can't tell if robert is just confused, or more likely, taking appreciation at the step of 'descent' =] .

 

[robert Ihnot]

Giz Z: I can't tell if robert is just confused, or more likely, taking appreciation at the step of 'descent=].

I was just hoping to go another step with the problem. INFINITE DESCENT? Yeah, just go look that up on Wikipedia, and by gosh! There is our problem.

 

[al-mahed]

I think this is a simple proof, using Fermat's theorem.

a^2 == -b^2 mod 3

if gcd(a,3)=1 ==> a^2 == 1 == -b^2 mod 3 ==> gcd(b,3)=1

but b^2 == -1 mod 3 is a contradiction ==> a and b must be divisible by 3

 

[robert Ihnot]

I was just trying to point out, as a hint, that the problem could be expanded to prove it had no non-trivial solutions. That is:

a^2+b^2 not equal to 3(s^2+t^2) unless a=b=c=d=0.


That was the intent. I guess it was not clear.

 

[Gib Z]

I think this is a simple proof, using Fermat's theorem.

When one uses a result much stronger then what is to be proven, It's sort of like using the mean value theorem to prove Roelle's theorem. Just an opinion though.

 

[al-mahed]

I was just trying to help

When one uses a result much stronger then what is to be proven, It's sort of like using the mean value theorem to prove Roelle's theorem. Just an opinion though.