Amazing! However i have a question if anyone could answer. I was able to follow most of it, but didn't understand how he chose the 2 close perfect squares, around 4:50 - 6:27. P.s tips about how to improve would also help :D
I think the strategy is to look at the higher order terms, since there is a^4 - 6a^2, he looked for (a^2 - 3)^2, then you can look for which values of a the linear term and the constant are higher or lower than each other, that is my take at least.
Radic has to be a square dictated by the 2 Pn terms of highest power. Its a bounding trick he often uses when Pn w/o sq repr. has to be square. Bound it to 2-3 values or soln doesnt exist. Look for his n^4 problems ...theres like 3 of them
To make you feel better this is standard olympic math technique. No normal kids can think how to do this on the spot for the first time. They learnt and practice and practice ... . If you can do this technique invented on your own , then you are a genius .
It's a short exercise to see that p = 19 produces 343, which is 7^3. Once again we have a problem whose only solution can be found by very simple mathematics. I still enjoyed the robust solution.
How do we know that if p divides (x²+x+1), then a factor of p-1 doesn't also divide (x²+x+1) which would mean (x-1) doesn't have to be divisible by p-1?
Always the best math vids on UA-cam 👍
Amazing .. I love the way the logic flows.
This is art. 🥰
Wowow... Nice exercise
🔥🔥
Why do questions like the shape cube instead of x³?
How would you choose the squares if you got something like m^4 + 3m^3 + ... with the cubic term
I suppose that 1 isn't counted as a prime. :-)
the definition of prime is based on non trivial divisibility/factorization, so unitary elements are never primes.
Amazing! However i have a question if anyone could answer. I was able to follow most of it, but didn't understand how he chose the 2 close perfect squares, around 4:50 - 6:27.
P.s tips about how to improve would also help :D
I think the strategy is to look at the higher order terms, since there is a^4 - 6a^2, he looked for (a^2 - 3)^2, then you can look for which values of a the linear term and the constant are higher or lower than each other, that is my take at least.
Radic has to be a square dictated by the 2 Pn terms of highest power.
Its a bounding trick he often uses when Pn w/o sq repr. has to be square. Bound it to 2-3 values or soln doesnt exist. Look for his n^4 problems ...theres like 3 of them
To make you feel better this is standard olympic math technique. No normal kids can think how to do this on the spot for the first time. They learnt and practice and practice ... . If you can do this technique invented on your own , then you are a genius .
So how about when is p^2-p-1 a cube? There's at least one obvious solution.
It's a short exercise to see that p = 19 produces 343, which is 7^3. Once again we have a problem whose only solution can be found by very simple mathematics. I still enjoyed the robust solution.
How do we know that if p divides (x²+x+1), then a factor of p-1 doesn't also divide (x²+x+1) which would mean (x-1) doesn't have to be divisible by p-1?
Because p-1 = (x^2+x+1) * ((x-1)/p)
Since (x-1)/p is an integer, x^2+x+1 divides p-1
p=1 doesn't work? why?
1 is not prime.
You ment why not x=1
Well p^2-p is always >0
N9
asnwer=1 os isit 🤣🤣🤣🤣