It seems to be much more trivial when you consider mod 6. Consider p, q, r ≥ 7. Then p, q, r = 1 or 5 (mod 6) p², q², r² = 1 or 25 = 1 (mod 6) So p² + q² + r² = 3 (mod 6) But then p³ = 1 or 125 (mod 6) = 1 or 5 (mod 6) ≠ 3 Hence no solution. We simply check cases p = 2, 3, 5 to get (p, q, r) = (3, 3, 3).
Nice video as always!👍 A long time ago you mentioned that you use (smoothdraw), do you still use it?, may I know what is your writing device? Your explanation always great and organised!👍
The case p^2; q^2; r^2 = 1 mod 3 doesn't lead to the solution with p=q=r=3. Is actualy simpler, and doesn't give any solution. The assumption p^2; q^2; r^2 = 1 mod 3 means that none of the primes is equal to 3 (otherwise at least one of them will be =0 mod 3, and so its square). With that assumption we find p^3=0 mod 3, that leads to p=3, giving us a contradiction. So the assumption p^2; q^2; r^2 = 1 mod 3 cannot be true, and this means that in any solution (p;q;r) at least 1 of them must be 3. From there the calculations are basicaly the same
That happened to me. The problem is that p^3 == 3 either if all variables are 3 or all are higher primes. But that's impossible p can't be both 3 and a higher prime.
I was initially confused how you know (p-1)/2 is odd at 7:07
You might want to mention it's because p = 4n+3 so (p-1)/2 = 2n+1.
Ty
Basic knowledge on topics 1. Quadratic residue 2. Quadratic reciprocity 3. sums of square from Number theory . will help solve this question.
what about the case if p=r≠3
It seems to be much more trivial when you consider mod 6.
Consider p, q, r ≥ 7. Then p, q, r = 1 or 5 (mod 6)
p², q², r² = 1 or 25 = 1 (mod 6)
So p² + q² + r² = 3 (mod 6)
But then p³ = 1 or 125 (mod 6) = 1 or 5 (mod 6) ≠ 3
Hence no solution. We simply check cases p = 2, 3, 5 to get (p, q, r) = (3, 3, 3).
Nice video as always!👍
A long time ago you mentioned that you use (smoothdraw), do you still use it?, may I know what is your writing device?
Your explanation always great and organised!👍
The case p^2; q^2; r^2 = 1 mod 3 doesn't lead to the solution with p=q=r=3. Is actualy simpler, and doesn't give any solution. The assumption p^2; q^2; r^2 = 1 mod 3 means that none of the primes is equal to 3 (otherwise at least one of them will be =0 mod 3, and so its square). With that assumption we find p^3=0 mod 3, that leads to p=3, giving us a contradiction. So the assumption p^2; q^2; r^2 = 1 mod 3 cannot be true, and this means that in any solution (p;q;r) at least 1 of them must be 3.
From there the calculations are basicaly the same
I don't get why you redid 3 again
Squares of prime 1 mod 6
x^n + (2+x)^n +(2-x)^n = 0 , x,n are integers
What?
@@iainfulton3781 asking whether anyone could solve it, but forgot to ask xd
You already solved for q and r = 3
That happened to me. The problem is that p^3 == 3 either if all variables are 3 or all are higher primes. But that's impossible p can't be both 3 and a higher prime.
Disappointing - yet again, the only solution is a glaringly obvious one!
you can apply a lemma, "If p is a 4k+3 prime and p|a^2+b^2, p|a and p|b".