A challenging divisibility problem!
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- čas přidán 20. 04. 2021
- We look at a divisibility problem that involves several number theory tricks.
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That first chart has an error... for n = 4 and m = 2 we have m^2 + 9 = 4 + 9 = 13, not 15. m = 6 does work, though, since 6^2 + 9 = 45 and that is a number divisible by 15 (45 = 3*15).
Nice try to catch us with 2²+9=15 but it's actually equal to 13 ;)
I thought that proving that p doesn't divide m^2+9 could be checked in a more satisfying way by considering quadratic residues. Clearly p divides m^2+9 iff m^2+9 ≡ 0 mod p. Then m^2 ≡ -9 mod p. In other words -9 is a quadratic residue mod p. So we need only compute the Legendre symbol L(-9,p). Now L(-9,p) = L(-1,p) L(9,p) = L(-1,p) L(3,p)^2. So if p ≠ 3, then L(3,p)^2 = 1, since L(3,p) is in {-1,1}. Therefore if p ≠ 3, L(-9,p) = L(-1,p). And L(-1,p) = 1 iff p ≡ 1 mod 4. So if p ≠ 3 and p ≡ 3 mod 4, then we have L(-9,p) = -1, so -9 is a quadratic non-residue mod p, so p doesn't divide m^2+9.
Astonishing that a man that knows all this number theory states 4+9=15 🤣🤣🤣
if m=2 it becomes 2^2 + 9 = 13 and not 15
Breach, broach, tomato, tomato.
Just to clarify: let p be a prime number congruent to 3 mod 4. Then p= 4k+3 for some integer k. Then m^2=-9 mod p and m^(2*(2k+1)) = - 3^(2*(2k+1)) mod p so m^(4k+2) = -3^(4k+2) mod p but from Fermat's Little Therom we know that 1=m^(4k+2) = -3^(4k+2) = - 1, so 2=0 mod p which is impossible
2^2 + 9 is not equal to 15, also 2^5 -1 = 32-1 = 31 so not 33 michael
There is a mistake in the chart m=2 is not a solution bc m²+9=13 not 15
2^2+9 doesn’t equal 15
Nice solution, this problem is from the 1998 shortlist N5, and I think that it is somekind classical.
At
Finally, some high quality problem!!!! thanks :D
Another highly interesting, educational and enjoyable video! How on earth did you manage to store up enough videos while your setup is being worked on? Are the outdoor/ whiteboard videos still ‘live’?
17:07
Best explanation ever
Nice video, learning number theory rn :D
Wrt why it possibly was not chosen as an actual IMO task:
OPEN PROBLEM : I often found interesting problems but because I haven’t found solutions, I’ve ditched them... until today. From now I will post them as open problems and I will not provide solutions or answers.
For the application of CRT you need to say the divisors are coprime. They clearly are in this case but would be nice to mention it