Find the max value for n

[Steven G]

Suppose n and sqrt(n^2 + 204n) are both positive integers. Find the maximum value of n.

 

[mario99]

I guess that you didn't read the posting guideline for this forum. Please do that and then post a comment according to does rules.

What have you done so far? Please, show your work so that we can know exactly at where you are stuck!

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[Otis]

show your work so that we can know exactly at where you are stuck

I'm not convinced that Steven is stuck. This thread is on the Odds&Ends board, so I take it as a math puzzle posted for recreational consideration by other members. (I've found the answer, but I cannot yet justify it. Still pondering…).
[imath]\;[/imath]

 

[blamocur]

Hint:

Spoiler

use the Pythagorean Triples.

 

[Steven G]

Hint:

Spoiler

use the Pythagorean Triples.

At some point I'd like to see how you would solve this one with your hint.

 

[Steven G]

I'm not convinced that Steven is stuck. This thread is on the Odds&Ends board, so I take it as a math puzzle posted for recreational consideration by other members. (I've found the answer, but I cannot yet justify it. Still pondering…).
[imath]\;[/imath]

What's your answer?

 

[blamocur]

At some point I'd like to see how you would solve this one with your hint.

It is not too elegant, but looks clean to me:

Spoiler: Solution

If [imath]n^2 + 204n = m^2[/imath] then
[math](n+102)^2 = m^2 + 102^2[/math]Thus for natural [imath]k,u,v[/imath]:
[math]102 = 2kuv, \;\;\;\; m = k(v^2-u^2)[/math]We can safely assume that [imath]u\ < v[/imath] and get these 4 cases:
[math]\begin{array}{|c|c|c||c|c|}\hline k & u & v & m & n \\\hline\hline 1 & 1 & 51 & 2600 & 2500 \\\hline 1 & 3 & 17 & 280 & 196\\\hline 3 & 1 & 17 & 864 & 768\\\hline 17 & 1 & 3 & 136 & 68\\\hline \end{array}[/math]
Another variant:
[math]102 = k(v^2-u^2) \;\;\;\; m = 2kuv[/math]
does not produce any new solutions. Here [imath]v^2-u^2[/imath] must be odd or divisible by 4, but 102 is not divisible by 4, thus [imath]k = 2[/imath] and [imath]v^2-u^2 = (v-u)(v+u) = 51[/imath].

[math]\begin{array}{|c|c|c|c|c||c|c|}\hline k & v-u & v+u & u & v & m & n \\\hline\hline 2 & 1 & 51 & 25 & 26 & 2600 & 2500 \\\hline 2 & 3 & 17 & 7 & 10 & 280 & 196 \\\hline \end{array}[/math]

 

[Otis]

What's your answer?

Same as blamocur's.
[imath]\;[/imath]

 

[Steven G]

Yes, the answer is n= 2500