Відео

Thank you for the video. After subtracting q² on both sides we get 36q² = (p-q)(p+q-1) thus p - q > 0. (p+q-1)-(p-q)=2q-1 thus one factor is even and the other one is odd. We can check that q=2 doesn't work (the quadratic equation for p has no integral solution). As both p and q must be odd, we must have p-q=4, 12 or 36 then p+q-1 is 9q², 3q² or q² (note that p-q=kq would make q a prime factor of the distinct prime number p, which is a contradiction). The third case gives the solution: p=q+36 & 2q+36-1=q² then (q-1)²=6² and p=q+36=7+36.

Check the solution: (p, q) = (7, 43) p^2 + q = 7^2 + 43 = 49 + 43 = 92 37q^2 + p = 37(43^2) + 7 Is not equivalent.

Favourite math channel!

My method, before watching the video, and probably not as rigorous as the referees would like. Should have done this using more explicit modulo notation, but that's not how my brain worked. A) For that to be a natural number, the numerator must be divisible by 6, which means it must be both even and divisible by 3. B) Is it always even? Well, n must be either even or odd. If it's even, the sum is even because each term is multiplied by n at least once, and any number times an even will result in an even answer. If it's odd, then the first term will be even because of the 2 coefficient, but the second and third terms will both be odd because they will be the product of all odd numbers. An even plus an odd plus another odd will result in an even number though. So regardless of the value of n, the numerator will be even. C) Is it divisible by 3? n must either be divisible by 3, 1 greater than a multiple of 3, or 2 greater. (i.e. n = 3k or n = 3k + 1 or n = 3k + 2). D) The numerator can be factored in a useful way. 2n^3 + 9n^2 + 7n = n(2n^2 + 9n + 7) = n[(2n + 1)(n + 4) + 3]. (Yes, I factored it in a clumsy way, but it still worked, and this was how I solved it.) E) If n = 3k, then the product above is clearly divisible by 3 because it has a factor of n. If n = 3k + 1, then the first parenthetical term in the brackets is divisible by 3, because (2(3k + 1) + 1) = 6k + 3, and the addition of a 3 outside of that product will not change this. If n = 3k + 2, then the second parenthetical term in the brackets is divisible by 3, because (3k + 2 + 4) = 3k + 6, and again, the additional 3 changes nothing. F) This expression will produce an even result that is divisible by 3 for all natural n, so it will always result in a natural number.

Interesting problem solution!

Excellent explanation!

Beautiful challenge!

i thought of 8-0-0; 5-0-3; 5-3-0; 2-3-3; 2-5-1; 7-0-1; 7-1-0; 4-1-3; 4-4-0 but its not as fast... 9 steps

Just tilt the 8 liter one so that you can see when half of it has fallen out.

This is too easy

For added difficulty: Prove that this is the fastest way, as in the least amount of steps, to do this.