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.

pt-short-long-legs-1

 
x \; = \; 3

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

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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.

      • 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.

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