1. ## Diophantine equation

I think Diophantine equations are in Number Theory...

In any event. I am trying to solve $25 n^2 + (10 a - 264) n - (3a^2 - 4a + 52) - p^2 = 0$ for n and p, where a is a constant positive integer.

I wasn't able to figure out how to do it by hand and I couldn't find a way to get W|A to understand the "a".

Any help is appreciated.

-Dan

2. ## Re: Diophantine equation

If we let $m=25n+(5a-132)$,
the equation can be written as

$(m-5p)(m+5p)=4 \left(4681-355 a+25 a^2\right)$

This says $m-5p$ and $m+5p$ are divisors of $4 \left(4681-355 a+25 a^2\right)$
Given $a$, we can solve for $m$ and $p$ and then find $n$

This is not the best solution. Ideally we would like to find a formula for $n$ and $p$ in terms of $a$

3. ## Re: Diophantine equation

One explicit solution

$n=-182+14a-a^2$

$p=936-71 a+5 a^2$

4. ## Re: Diophantine equation

I had a thought last night after I posted this. As I was finishing off the typing I noticed that it didn't work if a is to be an integer. Here's the rundown of what I came up with. Maybe someone can fix the details.

My strategy was to complete the square on n, make a redefinition to eliminate n from the LHS. This gives a Pell-like equation.

What I did:
$25 n^2 + (10 a - 264) n - (3a^2 - 4a + 52) - p^2 = 0$

Completing the square gives:
$\left ( 5n - \frac{10a - 254}{10} \right ) ^2 - p^2 = f(a)$

where $f(a) = 3 a^2 - 4a + 52 + \left ( \frac{10a - 264}{10} \right ) ^2$

Letting $m = \frac{10 a - 264}{10}$ gives a condition on a: m must be an integer. But according to this m is not an integer for any a. So it doesn't work.

-Dan

5. ## Re: Diophantine equation

Originally Posted by topsquark
$25 n^2 + (10 a - 264) n - (3a^2 - 4a + 52)$
Above has to be a square...(since it equals p^2).
Keeping that in mind in a looper, I get numerous solutions; example:
(a,n,p)
1,15,42
2,10,2
5,-9,62
6,10,18
....
97,47,246

58 solutions keeping a from 1 to 99
and n from -99 to 99.

Did I commit an unforgivable goof?

6. ## Re: Diophantine equation

Originally Posted by DenisB
Above has to be a square...(since it equals p^2).
Keeping that in mind in a looper, I get numerous solutions; example:
(a,n,p)
1,15,42
2,10,2
5,-9,62
6,10,18
....
97,47,246

58 solutions keeping a from 1 to 99
and n from -99 to 99.

Did I commit an unforgivable goof?
Thanks. I was looking for something more analytical, though. Diophantine equations are far from my field of expertise. For all I know maybe it can't be done.

-Dan

7. ## Re: Diophantine equation

If we let $m=25n+(5a-132)$,
the diophantine equation can be rewritten as

$(m-5p)(m+5p)=4 \left(4681-355 a+25 a^2\right)$

Finding solutions involves finding divisors of the quantity

$u=4681-355a+25a^2$ for a given $a$

Look for pairs $\left\{f_1,f_2\right\}$ such that $f_1f_2=u$, $f_2\geq f_1$ and $f_1\equiv 4$ mod 5

Then let $m=f_1+f_2$

example: $a=1$

$\left\{f_1,f_2\right\}=\left\{-4351,-1\right\}$ giving $n=-169$

and

$\left\{f_1,f_2\right\}=\left\{19,229\right\}$ giving $n=15$

8. ## Re: Diophantine equation

Oooooooooooh! That's clever. That answers my question nicely.

Thanks to all!

-Dan