👍

Sekali - kali bahas osp ya

Iya juga, ❤

❤

Nice

Not so hard

Well presented

Amazing

Sekali sekali bahas pelatnas bang

Terlalu cepat pembahasannya

Nice!

I think you mean Z^+ in the statement of the problem in the beginning rather than R^+. Also this is not a proof, since you didn't account for p-6 and p+6 not being prime, and hence having some other combination of prime factors on the right equaling the two factors on the left. An analysis of the possibilities for prime factors mod 6 works though. Namely, p-6 = p+6 = p (mod 6), so there are two cases: First case: They're both = 3 (mod 6), in which case p=3 (mod 6). Since p is prime, this forces p=3, and we are in the case you presented. Second case: They're both = 1 or 5 (mod 6). If they are = 1 (mod 6), then they individually have an even number of prime factors congruent to 5 (mod 6). If they are = 5 (mod 6), then they individually have an odd number of prime factors congruent to 5 (mod 6). Either way, when you pool all the prime factors on the right, you will have an even number which are congruent to 5 (mod 6). Then, a regrouping of the factors into say k and m, will always have either an odd or an even number of factors = 5 (mod 6) in each grouping. In the first case, k=m=5 (mod 6), and their sum will be k+m = 10 = 4 (mod 6). In the second case, k=m=1 (mod 6) and their sum will be k+m = 2 (mod 6). However, when you consider the system of equations that arises from this grouping, it looks like: 3a+2b=k 3a-2b=m Adding, we have 6a=k+m. From above, we know k+m = 2 or 4 (mod 6). But 6a=0 (mod 6), so this is a contradiction. So this case cannot occur; we must have p=3 (mod 6).

nice

Akhirnya bahasa Indonesia yuhu

Mantap bang penjelasannya

24 = A^2 - n^2 = (A+n) (A-n) = 6 x 4 => A = 5 and n = 1 (for A,n to be integers, the factors need to be both even or both odd but not even-odd due to division by 2 unlike 8x3=24). 9 = n^2 - B^2 = (n+B)(n-B) = 3 x 3 => n = 3 and B = 0 OR 9x1 so n = 5 and B = 4 Only common positive integer solution is n = 5 to get A-B = 7 - 4 = 3

*Here are all 5 integer solutions for A* Check if x+y = 0 = a works. It does, so that's 1 answer. Then, assume non-zero and divide everywhere using (a^n + b^n) / (a+b) for odd n (1,3,5) formula, to get 1 = x^2 - xy + y^2 = x^4 + y^4 + (xy)^2 - xy (x^2 + y^2) = a/a = 1 a^2 = (x+y)^2 = x^2 + y^2 + 2 xy = 1 + xy + 2xy = 1 + 3xy, so xy = (a^2 - 1)/3, so x^2 + y^2 = 1 + xy = 1 + (a^2 - 1)/3 = (a^2+2)/3. Lastly, apply (x^2 + y^2 = (a^2+2)/3 AND xy = (a^2 - 1)/3 into the expansion above of (x^5+y^5)/(x+y) to get a quadratic equation in a^2 as follows, 1 = x^4 + y^4 + (xy)^2 - xy (x^2 + y^2) = (x^2 + y^2)^2 - (xy)^2 - xy (x^2+y^2) = common denominator 9, so = 1/9 [ (a^2+2)^2 - (a^2-1)^2 - (a^2-1)(a^2+2) ] So, (2a^2 + 1)(3) - (a^2)^2 + a^2 - 2) = 9 => (a^2)^2 - 5 a^2 + 4 = 0 => (a^2 - 1)(a^2 - 4) = 0 => a^2 = 1 or 4, so a = +/- 1 or +/- 2 *Finally* -2 <= a <= 2 (that would be all 5 integer solutions of a). [ x and y are also computable from a^2 if needed using xy = (a^2-1)/3 = 0 or 1 and x^2 + y^2 = (a^2+2)/3 = 1 or 2 ]

With large numbers, checking if it's prime or composite is so much easier than finding its factors. For example, is this number prime or composite ? *1180486325947394189842626845060356983501034478265470980260373* What are its prime factors if it is composite ?

EXCELLENCE

افضل حل

Soalnya settingan

It is too easy

Ada baiknya jika menyertakan sumber olim 🙏🏻🙏🏻

By geometry series

Im sleeping

Clear as crystal

Good work, Dude!

Simpel

good

Good jobb ❤

Simpel sekali

How can you find the solution ?

Nato Should removed there occupation force from Serbia or Ukraine should stop wanting to occuping 4 inderpent country.

Why can’t x=y=1 so a=2 or x=y=-1 so a=-2?

Soal OSK SMA 2023 nihh

Well solved

How did you get (2n-1)!!

Ajarin matematika pak

Beautiful proof❤

👍

Good❤

Gak ada yg lebih easy lagi😊😊