## Pythagorean triangles – shortest leg, longest leg

Find all values of a positive integer constant   $n$   such that

$x^2 \; + \; n \,x$      and      $n \,x \; + \; n$

describe the two legs of a Pythagorean triangle for every positive integer value of   $x$.

Let   n = 2

Then,

$a \; = \; x^2 \; + \; 2 \, x$
$b \; = \; 2 \, x \; + \; 2$
$c \; = \; x^2 \; + \; 2 \, x \; + \; 2$

$(x^2 + 2 \, x)^2 \; + \; (2 \, x + 2)^2 \; = \; (x^2 + 2 \, x + 2)^2$

True for all   $x$

Is   n = 2   the only solution?

$x \; = \; 1$

$4 \, n^2 \; + \; (n+1)^2 \; = \; 5 \, n^2 \; + \; 2 \, n \; + \; 1$

$5 \, n^2 \; + \; 2 \, n \; + \; 1 \; = \; y^2$

getting a Diophantine equation. There are infinitely many solutions.

$x \; = \; 3$

$16 \, n^2 \; + \; (3 \, n + 9)^2 \; = \; y^2$   —->   $n = 2, \; y = 17$

math grad - Interest: Number theory
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### 3 Responses to Pythagorean triangles – shortest leg, longest leg

1. Paul says:

n must be just 2 because when x=3 we get

(3^2 + n 3)^2 + (n 3 + n)^2 which simplifies to 81+54 n+25 n^2

and the only solution when that equals a square is when n = 2, the square is then 17.

A test shows this is true for all x up to a lot.

Paul.

• benvitalis says:

Check out my post

• Paul says:

“Is n = 2 the only solution?” For all x..

Yes it is .

There are an infinite numbers of solutions for n when x = 1 and 2 and the only common value for n is 2, but when x = 3 there is only 1 solution for n as I mentioned above and again the value is 2, so there can never be another n beyond x = 2 that is common.

The first few solutions for n when x = 1 are

{2, 15, 104, 714, 4895, 33552, 229970, 1576239, 10803704, 74049690} which has a linear recurrence of {8, -8, 1}

and when x = 2 the first few values of n are

{2, 16, 102, 320, 1978, 12210, 38150, 235438, 1452960, 4539602}

linear recurrence of {1, 0, 119, -119, 0, -1 ,1}

Paul.