Find the maximum value of n

This problem was a mathematics competition. The strategy is to complete thew square and use the properties of integers to find the maximum of of n

[KevinTheall]

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.

[avupatimunna]

Excellent explanation on how to approach such questions

[PavelSVIN]

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

[williamperez-hernandez]

Quickest to solve for n was using: n + 102 = k. We have k = r +2 = 2602. So n= 2602-102, n=2500.

[pietergeerkens]

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

[freddyalvaradamaranon]

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. ❤😊❤😊.

[aCalifornian]

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

[stevenwilson]

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.

[mab]

Hey man. Your solutions are just fantastic. Thanks.

[GameX]

Amazing question sir.
Thanks for sharing this question
And for the amazing explanation keep up the good work
Love from India ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

[mahan]

Great job!
You could also have solved the equation for k (instead of r) and directly find n, as k is equal to n+102

[cesarjmachado]

Thank you. All possible positive solutions for n are 68, 196, 768, and 2500. Plugging factors for the equations: (1, 51^2), (3, 3*17^2), (17, 17*3^2), and (3^2, 17^2).

[skwbusaidi]

Excellent. It would have been eaiser to find k instead of r
k= 51^2+1
k= n+102
n=51^2-101
n= 51^2-1-100
n=(51-1)(51+1) -100
n=50×52 -100
n=2600-100=2500

[nanashi__]

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

[florianbuerzle]

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.

[nothingbutmathproofs]

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!

[childrenofkoris]

this is crazy.. please more more more 🤓🤓🤓🤓🤓🤓🤓

[artandata]

great !! thanks for this video, master !

[hr]

I like your idea of ​​adding a passage from the Bible at the end of the video. 😀

[secret]

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)?