A Fast and Simple Equation in National Maths Olympiad | All-Soviet Union Mathematics Olympiad 1981

[tianqilong]

Although we cannot directly assume they are natural numbers, but we can prove that x, y >0 has to be the case, here is my thought:

if x>0, y<0, then xy<0, so x^3 < 64, x<4, so we still just need to consider x=1, 2, 3 as what has been discussed in the vid.

if x<0, y>0, then LHS is negative, so |xy|>61, but then the maximum of LHS occurs at -x=y=8, which is already much smaller than xy+61=-3, and LHS decreases much faster than RHS as |x| and |y| gets larger, so this case can just be ignored as well.

if x, y<0, it is just the exact same as the case where x, y>0, we just need to convert the y to -x and x to - y from case[x, y>0].

Proved, only x, y>0 matters.

[smallyellow]

(x, y)=(-5, -6) also satisfies the formula, then (x, y)=(-5, -6) should be included in the answer.

[parthibhayat]

Since imo is going on, I think it would be nice if you looked onto the imo problems anytime later

[cspitney]

Hi. I'm not clear how you assumed x, y are Natural?

[parthibhayat]

Everytime I see the videos, I feel like I am getting better at these Diophantine equations, if that's what they are called

[riadsouissi]

For the general case (a, b) can be negative, we can show that if 3a-1 divides 61-a^3 this implies 3a-1 divides 3.61-a^2 then divides 9.61-a then divides 27.61-1 = 1646 = 2.823 (823 is prime). From there, 3a-1 can take the following values (1, -1, 2, -2, 823, -823, 1646, - 1646). For large values of a, the calculations become a bit annoying without a calculator but it is doable.

[petersievert]

1:13
But x and y are allowed to be negative after all, aren't they? So it is not that easy to assume a, b > 0 .
Or is the problem in the thumbnail?

[mathcanbeeasy]

Integers doesn't mean natural. You did not solved the original problem, but a different one.! 🤔

[forte]

61 = 25 + 36
Дальше можно догадаться что одна из переменных будет равняться 6 или 5

[oguzhanozdogan]

Good jop.

(20.5^n-2)/(3^n+47) find n that make integers
n (positive integers)

[caesar_cipher]

The creator of this channel should himself start thinking critically - incomplete definition if question, missed logic - doesnot learn anything from past mistakes in other videos

[serpentdenuit]

Other way : we assume that x, y are natural (see the message of 龙天骐 to show why we can). Then x^3-y^3>0 so x>y and we have y<=x-1. But we want :
so x^3<=(x-1)^3+x(x-1)+61. This leads to -2x^2+2x+60>=0, and since the roots are -5 and 6, we have x<=6.

If x=6 we have y^3+6y=155 so y=5. If x=5 we find y^3+5y=64 which has no solution, and so on. Finally the only solution for x, y natural is (6;5) (and the 2 solutions of the equation are (6;5) and (-5;-6)).

[tianqilong]

x=5, y=4 can also immediately fit just by looking at the thumbnail