Poland Mathematical Olympiad 2024 | The Secret to Mastering Radical Equation Fast

[@JLvatron]

How do you conclude 3:55 ? Had the term been something like sqrt(12a), we wouldn't conclude a=12*x^2; as 'a' could be 3, so sqrt(12*3)= sqrt(36) = 6.

[@ThePhantomoftheMath]

Hi, and thanks for watching! That couldn’t have happened because, as you can see at 2:48, we first simplified 2024 as the product of prime numbers. So, it’s impossible for the term √(12a) to appear later, as it would have already been simplified as 2 * 2 * 3. The number 2 would have been moved in front of the square root (as an integer), leaving us with √(3a). In this case, a=3*x². Hope that makes sense.

[@JLvatron]

@@ThePhantomoftheMath That does make sense. Thank you!

[@michaeldoerr5810]

The video explanation is much MUCH better than the comments!!!

[@waheisel]

I had a similar though possibly simpler and less rigorous explanation: The only integer that can be removed from the radicand in sqrt2024 is 2. The only pair of non negative integers that add to 2 are 0,2 and 1,1. So by the distributive property all solutions for a and b are 0,2024 (and vice versa) and 504,504.

[@Antony_V]

I have the impression that there is something wrong: at 3:57 √(506*a) must be an integer, ok, so you wrote a=506*(x^2). For me it would be a*506=x^2. Since 506 is the product of three prime numbers the smallest square under the root is 506^2. That gives a=506. The same for y, at 5:29, y=506. The others 2 solutions (0,2024) and (2024,0) obviously don't need any calculus.

[@denisrenaldo3506]

I think that you can easily deduct that sqrt(a)+sqrt(b)=2sqrt(506) and from that specific equation deduct the panel of solutions [0;2024], [506;506] and [2024;0]. That’s it !

[@nobody08088]

I’d gotten to the first step of your solution myself before seeing your comment and was disappointed I didn’t get to solve it myself 😢

[@denisrenaldo3506]

@@nobody08088 as a and b are integers the sum of the radicals which is sqrt(2024) or 2sqrt(506) can only be split in two ways : 0+sqrt(2024) or sqrt(506)+sqrt(506) and from that point, you deduct the panel of solutions

[@RAG981]

It makes no sense to allow a zero solution for either a or b, since the solution becomes trivially that the other is 2024. Both being rt506 is the only non trivial solution.

[@oqardZ]

Hi! Might I respectfully ask who appointed you the arbiter of which solutions are or aren't allowed? (Edit: fixed a typo.)

[@ingiford175]

It said non-negative. Which zero qualifies as. And yes, it is a trivial solution.

[@pietergeerkens6324]

@@ingiford175 Exactly! The question is what it is, and must be answered as such for full marks. In this case, I might think that the trivial solutions are deliberately included to allow weaker students to earn part marks.

[@KipIngram]

Well, obviously a = 2024, b = 0 and a = 0, b = 2024 are solutions. To look for other solutions, allow q^2 to be some unspecified rational number such that b = q*a, and substitute: sqrt(a) + sqrt(q^2*a) = sqrt(2024) sqrt(a)*(1+q) = sqrt(2024) a*(1+q)^2 = 2024 a = 2024/(1+q)^2 Here a must be an integer, so (1+q)^2 must be a factor of 2024 - an integer. Note that (1+q)^2 = 1 + 2*q + q^2. We already have q^2 rational, so here we see that q must also be rational. Allowed values for (1+q)^2 are the factors of 2024 = 2^3 * 11* 23. The only factor we can build from these that has a rational square root is (1+q)^2 = 4, so 1+q = 2 or q = 1. So, a and b are equal and a*(1+1)^2 = 2024 4*a = 2024 a = 506 Our answer is a = b = 506. Q.E.D.