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!

READ BEFORE POSTING

Welcome to FreeMathHelp.com! Please take the time to read the following before you make your first post. It will help you to get your math questions answered promptly and in the most helpful manner. A summary is available here, but please read the complete guidelines below at your earliest...

www.freemathhelp.com

 

[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…).
$\;$

 

[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…).
$\;$

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

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

 

[Otis]

What's your answer?

Same as blamocur's.
$\;$

 

[Steven G]

Yes, the answer is n= 2500