Alternatively, this problem can be solved using the Pythagorian triple parameterization: Let a,b,c be positive integers. For a² + b² = c² Set a = m² - l², b = 2ml and c = m² + l² where m>l are chosen positive integers. Apply this to the equation k² + (102)² = (n +102)² where n and k are positive integers to be determined. Set 2ml = 102 to obtain ordered pairs of positive integers (m,l) element of {(51,1), (17,3)}. Using Pythagorian parameterization, the pair (17,3) leads to k = 280 and n = 196. The first pair (51,1) leads to k = 2600 and n =2500, the desired result.
as opposed to the factoring required to get 2500 and 2704? just multiply 51*51 on paper then subtract 101. gives the correct answer of 2500 without nearly as much fuss
There is a shorter and simpler way. We know there is a perfect square under the square root. It means n²+204n=(n+k)² where k - some positive integer. From here we get n=k²/(204-2k). 204-2k is even as it can be presented as 2*(102-k), where k is an integer. Thus k² should be even as n is an integer, so k is even. Maximum n is when an expression 204-2k is minimal and positive (remember n is positive and k is positive and even) - it is when k=100. n=100²/(204-2*100)=2500
Once we get to n=k^2/(2(102-k)), we can set three conditions for k to help us find the value of k. 1. K is an even number (k^2=2n(102-k), making k^2 even. Because of this k must be even) 2. 102-k cannot result in a negative value, as it doesn’t make n a positive integer. This shows that k must be less than 102. 3. The whole point of this is to find the max value of n. If we are trying to find the max value of n, k must also be a max value. Putting these conditions together, k=100 is the only logical answer. If i made any mistakes, please let me know.
A much easier solution: r = sqrt(n^2 + 204 * n) = sqrt(n) * sqrt(n + 204) n is a square, also (n + 204) is a square. Let k = sqrt(n), j = sqrt(n + 204) k is maximized when the gap between k and j is the smallest possible Since 204 is even, the gap has to be even. Hence j should be equal to k + 2 to maximize k. j^2 - k^2 = 204 (k + 2)^2 - k^2 = 204 k^2 + 4k + 4 - k^2 = 204 4k + 4 = 204 k = 50 hence n = k^2 = 2500
Neither n nor n+204 need to be square. That's incorrect reasoning. There are 3 values of n that this works for: n=196, 768 and 2500. 2500 is the maximum, and it is square, but the other two are not and they work just as well.
@@danfoster8219To compute the maximum value of n, it has to be a perfect square. Suppose n is not a perfect square, it can only be a product of a perfect square and a factor of 204. However, that will never give us the maximum possible value of n. You suggested 768. That is 256 * 3. 3 is a factor of 204. Factor out 3 from all the numbers. You get 256 and 256 + 68 = 324. sqrt(256) is 16 and sqrt(324) is 18. For n = 768, we have n + 204 = 972. sqrt(n(n + 204)) = sqrt(768 * 972) = sqrt(256 * 3 * 324 * 3) = 16 * 3 * 18. You see 3, the selected factor of 204, showing up in the final answer. By the same token, we can calculate many values for n by exploiting the factors. None of these will give us the maximum value of n.
Nice problem!. If you know the perfect squares up to 24², this mental math trick quickly gives you the perfect squares from 26² to 74², for 1 < n < 25: (50 ± n)² = 50² ± 2⋅50⋅n + n² = 2500 ± 100n + n². For example: 37² = (50 - 13)² = 2500 - 1300 + 169 = 1200 + 169 = 1369, 73² = (50 + 23)² = 2500 + 2300 + 529 = 4929, and 51² = 2500 + 100 + 1 = 2601. Similarly, for the same range of n, we get perfect squares up to 124² as (100 ± n)² = 100² ± 2⋅100⋅n + n² = 10000 ± 200n + n²
Profesor Tambuwal muchas gracias por compartir tan buen video, explicando con detalle el procedimiento. Mi hija y mi persona estamos muy agradecidos de nuevo con su bella persona. ❤😊❤😊.
This was utterly brilliant. I immediately recognized that 0 was a solution. That was super easy to find. Of course that's not a positive solution so that doesn't really help. To find this max value tho, that was next level. Bravo.
Awesome video 😀 And great trick dividing two times by 2… if I had realized that, it would have saved me a lot of time 😂 My solution was similar, however I did not introduce a new variable for the binomial expression. Using your variables, I wrote: (n + 102)² - r² = 102² (n + r + 102)(n - r + 102) = 102² And now, the difference of two factors of 102² must be as large as possible and each factor must be even, so: 102² = 2 ∙ (2 ∙ 3² ∙ 17²) = 2 ∙ 5202 Now we get the following system of equations: n + r + 102 = 5202 n - r + 102 = 2 Adding these equations and solving for n: 2n + 204 = 5204 2n = 5000 n = 2500.
Nicely done. Not only could I not have solved this one easily but I would have insisted that you could find larger and larger values for n such that r would be an integer. Very surprising result. Thanks!
Is this meant to be without a calculator? We would need to factor or do the quadratic formula for (n^2)+204n-(2600^2) by hand (which, I suppose would be less problematic for Olympiad level students than it would be for me)?
@@PrimeNewtons That’s crazy. I doubt I could do it without a calculator, but I guess that’s why I didn’t do Olympiads! Thanks for your responses and the interesting video
Maybe, the intention was to instead of finding the value of 'r' and solving a quadratic, find the 'k', which is 2602, and by definition, k is n+102 so you would know n=2500
I solved it with a hyperbola: √(x²+204x) = y where this "y" should be a positive int. ↓ x²+204x = y² ↓ (x+102)² - y² = 102² and one of the asymptotes of this parabola is y=x+102 Let the first quadrant part of the parabola as a function "f". Domain: (0,∞) Range: (f(0),∞) While n is positive int, f(n) also should be a positive int Also, f(n) is always smaller than n+102 (Think about the asymptote i said) [1st attempt: When f(n)=n+101] Put x=n and y=n+101 (x+102)² - y² = 102² (n+102)² - (n+101)² = 102² ... (skip) ... n=5100.5 (sus 🤔) [2nd attempt: When f(n)=n+100] Put x=n and y=n+100 (x+102)² - y² = 102² (n+102)² - (n+100)² = 102² ... (skip) ... n=2500 (oh ye) Answer: 2500
This is absolutely crazy problem! It doesn't look that hard, but I wasn't even close to solving it despite of trying hard. I start solving it in ways that makes sense to me, but very fast I find myself in dead ends that don't make sense at all. So n²+204n must be an integer squared. Extending the square n² with "sidebands of width k" leads to n² + 2*k + k². We get 204n = 2*k + k². Since 204n is always even, k must be even too: k = 2b, were b is integer. (n+2b)² = n²+ 4nb + 4b² = n²+204n => n(b) = b² /(51-b). This never gives integers!! What went wrong?
Bro, you did 99%. Please see above my comment where I applied the same logic and finally got the answer. In your case b should be equal to 50 - in this case b2/(51-b) is max. And (surprise!) n=50^2/(51-50)=2500
@@PavelSVIN Yes, you are right! I was very close to the solution! Somehow I didn't see it and instead I thought I was in a dead end. On a good day I would have solved this.
if i am talking higher level maths in grade 10 next year, what topics may i encounter? i imagine for certain matrices, working with imaginary numbers or exponential equations. any other ideas?
would i not be easier to use quadratic formula rather than factoring the last equation n=[-204+-sqrt(204^2+5200^2)]/2 yes it will require some work but all seems rather straightforward compared to factoring
What I dont get is how did we even arrive at an answer. Given the initial conditions, what stops us from determining it's infinity? sqrt(inf^2+204inf) feels like a perfectly valid maximum (i know infinity isnt really a positive integer but you get the point). What actually causes a maximum to exist?
Maximum exist in the integer numbers. In the video he said let our square root be “r”, then raised both sides by square, added 102^2 and he got r^2 + 102^2 = k^2, where r is integer by condition and k we made be integer. And we got the Diophantine equestion (k-r)(k+r) = 102^2. And only cause k and r are integer, we have countable amount of solutions. And from these solutions we found which have the biggest value of n.
Think about it this way: the distance between squares increases when the numbers increase. Solving r^2=n^2+204n for n we get n=-102+sqrt(r^2+102^2) => k^2=r^2+102^2 k^2-r^2=102^2 There will be an upper limit above which the difference between the squares of any given integers k, r is greater than 102^2 Let k=r+m to get 2mr+m^2=102^2 r=(102^2-m^2)/(2m) and because r>0 m is limited.
Essentially, it is because the expression n² + 204n generates only a finite amount of perfect squares. So we could instead solve the problem: „Find all positive integers n for which n² + 204n is a perfect square.“ Using the same method shown in the video, we find that there are only four positive integer solutions: n = 68, 196, 768, 2500 of which n = 2500 is the largest.
Thanks for the replies, guys, but this still doesn't answer my question. Why does the maximum exist in the first place? We reached this because we manipulated the equations. However, with the equations we started with, why does a maximum even exist in the first place? I get how we reached this answer; however, this answer exists BECAUSE of the manipulations we did, which introduced a new variable with different constraints. However, for the original problem that is exclusively in n, an upper bound should theoretically not exist, no? The value of n can just increase up to infinity with no issues at all.
i saw this question and tried solving it this way but I didnt get the correct answer, please tell me where I went wrong. n(n+204) is a square number. so n + 204 = (k^2)n where k is some positive int. n = 204/(k^2 - 1) we need the max value of n. So k^2 - 1 should be minimum but also be a factor of 204 so k = 2 n = 204/3 n = 68 I think there is some mistake in my logic please help me here (I'm not very good at math sorry)
We know there is a perfect square under the square root. It means n^2+204*n=(n+k)^2 where k - some positive integer. From here we get n=k^2/(204-2*k). 204-2*k is even as it can be presented as 2*(102-k), where k is an integer. Thus k^2 is even as n is an integer, so k is even. Maximum n is when an expression 204-2*k is minimal and positive (remember n is positive and k is positive and even) - it is when k=100. n=100^2/(204-2*100)=2500
he looked at possible factors to find ones with a difference of 204. time consuming and difficult. willing to do some long multiplication would have just used quadratic formula. of course if he hadn't chosen to solve for r and instead solved for k we could have avoided the whole thing. we knew k=n+102. if we solved for k instead of r we would have got k=n+102=51^2+1 so n=51^2-101
