A Nice Problem From Sweden | Viewer Suggested 😊

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@GLRAVGJSKSG-lr8bx 10 ай бұрын

At 6:00 you could also use Fermat's Little Theorem: If q does not devide a, then a^(p-1) is congruent to 1modp In that case a=q and p=3. Because q is prime and q>3, obviously 3 does not divide 3. Hence, by F.L.T the relation: q^2 is congruent to 1mod3 or equivalent 3|q^2-1

@NirDagan 11 ай бұрын

Theorem: if a natural number q is prime and q>3, then q^2-1 is divisible by 3. Proof: q^2-1 =(q-1)(q+1). One of q-1 q q+1 is divisible by 3. so since q is prime and q>3 it must be that either q-1 or q+1 is divisible by 3. QED

@SyberMath 11 ай бұрын

Number Theory is awesome! 😍

@leif1075 11 ай бұрын

@SyberMath there is an issue here..I can easily write 2^q as 1 mod 3 NOT negative 1 since for example 256 divided by 3 has remainder 1 mod 3 not negative 1..so many ppl are not going to think of negative 1..see what I mean and why math can be arbitrary crap sometimes?? Surely you agree and hope you can PLEASE respond when you can.

@lettucehelper 11 ай бұрын

@@leif1075but q is prime and q>3. 256 = 2^8 so it does not fit the criteria. All primes greater than 3 are odd, and all odd powers of 2 are congruent with -1 mod 3.

@bjornfeuerbacher5514 10 ай бұрын

Calling this a "theorem" is a bit overblown. ;) "Lemma" or "proposition" would be more appropriate.

@KeioAccelerg 11 ай бұрын

In Japan, Kyoto Univerity has this quiz in 2016

@SyberMath 11 ай бұрын

Any links plz? 😍

@mcwulf25 11 ай бұрын

Nice problem and solution. Of course when you get to r==q^2-1 you also have r == (q+1)(q-1) (mod 3) We know 3 doesn't divide q if q € P >3, so 3 must divide either q-1 or q+1 as we are considering 3 consecutive numbers.

@pieintegration4083 10 ай бұрын

Nice!

@SyberMath 10 ай бұрын

Thanks!

@thomaslangbein297 11 ай бұрын

Little Fermat makes it even easier

@h2omathgaming453 11 ай бұрын

Thank you for bringing up Number Theory, one of my favorite Mathematics Topics! However, I think I know an easier way to prove why there are no other solutions beside q = 3. If q is a prime not 3 (and not 2 either), then q will be an odd prime not divisible by 3. Therefore, 2^q will always have a remainder of 2 when divided by 3, as it is equivalent to the odd power of -1, resulting -1 = 2 (mod 3). Meanwhile, by quadratic residue theory, q^2 will always have a remainder of 1 when divided by 3, since q is not a multiple of 3. Therefore, if q > 3, then 2^q + q^2 = 2 + 1 (mod 3) = 0 (mod 3). In other words, p^q + q^p is always a multiple of 3 if p = 2 and q > 3.

@mathswan1607 11 ай бұрын

Better to write q=6n+1 or q=6n-1 Since q=3n+/-1 may lead to q to be even when n is odd

@Noobish_Monk 11 ай бұрын

p^q + q^p is even if p and q are odd, so let q = 2. 2^p + p^2 mod 3 is almost always 1 since p^2 is 1 if p > 3 and 2^p = 1 (mod 3) if p is odd. So any p > 3 isn't good. So we can then easily get that the answer is (2, 3) and (3, 2)

@遠傳五華 5 ай бұрын

Cool!

@Qermaq 11 ай бұрын

A very clever problem. Most p^q are odd, so the sum would be even unless p or q = 2. So q = 2 and p is an odd prime. p^2 + 2^p = prime. Now other than 2 and 3, all primes are congruent to 1 or 5 mod 6, but all those other primes squared are 1 mod 6. 2^p is 4 mod 6. So the sum will never be a prime unless p=3. Then 4 + 3 = 1 mpd 6, and 17 happens to be prime. Yay!

@SyberMath 11 ай бұрын

😍

@Nikos_Iosifidis 11 ай бұрын

Very nice solution

@SyberMath 11 ай бұрын

Thanks

@FisicTrapella 11 ай бұрын

👌

@parveshsehgal901 11 ай бұрын

We can also take mod 6 I did it. By taking mod 6

@ひろ-j9s 11 ай бұрын

I am Japanese . This question was asked in the Japanese entrance exam. ( Kyoto University )

@SyberMath 11 ай бұрын

Interesting. I had seen some entrance exam problems in a math journal many years ago. The problems were pretty rigorous. Do you know any links to sites that has those problems or any books? Thanks in advance! ☺️

@gmncnr 11 ай бұрын

I knew this from michael penn

@tontonbeber4555 11 ай бұрын

This one is very easy ... 2^3+3^2 is the only solution, If both p,q are odd then sum is even not prime, so one must be even and 2 is only even prime If p odd not multiple of 3, q=2 then 2^p mod 3 = -1 and p^2 mod 3 = +1 so sum is multiple of 3, not prime.

@SyberMath 11 ай бұрын

Nice