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I note that the Diophantine equation, $x^2 + y^2 = z^2$, with $x, y, z \in \mathbb{N}$, has infinitely many solutions. Indeed, $(x, y, z) = (3,4,5)$ provides a solution, and for any $k \in \mathbb{N}$ : $(kx, ky, kz ) = (3k, 4k, 5k)$ provides a solution.

However, assuming $x, y, z \in \mathbb{N}$ with $x, y > 1$, is the same true for the Diophantine equations,

$x^2 +y^2 = z^2 + 1$,

$x^2 + y^2 = z^2 + 2$,

$x^2 + y^2 = z^2 + 3$

and more generally, for $x^2 + y^2 = z^2 + n$, for any $n \in \mathbb{N}$?

In particular, are there infinitely-many triples $(x, y, z) \in \mathbb{N}^3$ for which $x^2 + y^2 = z^2 + n$ is true for infinitely-many values of $n \in \mathbb{N}$?

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A related question that asked for something stronger (the density of the set of solutions) : math.stackexchange.com/questions/165698/…. The answer to your question seems to be that for any $n$ there will be infinitely many solutions to $x^2+y^2 = z^2+n$ –  mercio Feb 8 '13 at 14:24
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For $x^2+y^2=z^2+1$ note that $2x^2=z^2+1$ is known to have infinitely many solutions eg $(x,z)=(1,1),(5,7),(29,41) \dots$. –  Mark Bennet Feb 8 '13 at 14:25
    
Note also that $(2n+1)^2+(2n^2+2n-1)^2=[2n(n+1)]^2+2$ –  Mark Bennet Feb 8 '13 at 14:33

4 Answers 4

up vote 7 down vote accepted

$x^2+y^2 = z^2+n$ is equivalent to $x^2-n = (z-y)(z+y)$.

Any composite odd number can be written as $(z-y)(z+y)$ for some integers $z$ and $y$, so it is enough to show that $x^2-n$ contains infinitely many composite odd numbers.

If $n$ is odd, then you can simply pick $x = 2kn$ for any $k$. Then $x^2-n$ is a multiple of $n$ and odd, which gives you two integers $y$ and $z$ satisfying the equation. You end up with the family of solutions $(2kn, 2k^2n-(n+1)/2, 2k^2n-(n-1)/2)$

If $n$ is even, then you can simply pick $x=2k(n-1)+1$ for any $k$. Then $x^2-n \equiv 1^2-1 = 0 \pmod {n-1}$, and it is odd so again this gives you two integers $y$ and $z$ satisfying the equation. You end up with the family of solutions $(2k(n-1)+1,2k^2(n-1)+2k-n/2,2k^2(n-1)+2k-1+n/2)$

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In your solution for odd values of $n$, I think you must mean the triple $(2kn, 2k^2 n - (n + 1)/2, 2 k^2 n + (n - 1)/2$, else we do not get $x^2 + y^2 = z^2 + n$. –  Harry Williams Feb 10 '13 at 17:14
    
I am still not clear how you found $y$ and $z$. Is it just a parameterisation that happens to work, or is there some method in finding it? –  Harry Williams Feb 10 '13 at 17:18
    
@HarryWilliams : if $a$ and $b$ are odd numbers, then $ab = (z-y)(z+y)$ where $z=(a+b)/2$ and $y=(a-b)/2$. So for the case where $n$ is odd, I just wrote $x^2-n = n(4k^2n-1)$ and used that factorization to find some $y$ and $z$. And yes, I made a sign error. –  mercio Feb 13 '13 at 14:30
    
That's brilliant. Thank you very much. –  Harry Williams May 6 '13 at 23:28

EDIT: I can't quite parse the final question in the original post. However, given any integer $N,$ there are infinitely many triples $(x,y,z)$ that solve $$ x^2 + y^2 - z^2 = N^2 $$ using the process below. A "seed" triple may be taken with any $x=z, \; y = N.$ If we just take $\gcd(x,N) = 1$ as well, we get primitive triples.

ORIGINAL:Actually, the group of automorphs of $x^2 + y^2 - z^2$ is known. It is infinite, and generated by three matrices, along with negating any of $x,y,z$ if that is necessary, not sure. I am trying to find a question that shows the matrices, no luck so far. However, what is means is that, if there is a single solution to $x^2 + y^2 = z^2 + k$ for some integer $k,$ positive or negative or zero, then there are infinitely many, and we can travel among them by matrix multiplication of the column vectors $(x,y,z)^T.$

I think it was some question here or on MO more likely, about the structure of Pythagorean triples. Note that the expert on this is named Ian Agol. There is just a comment by him at this one:

http://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples

Here we go, http://en.wikipedia.org/wiki/Tree_of_Pythagorean_triples
these matrices due to F. J. M. Barning (1963)

$$ A = \; \left( \begin{array}{rrr} 1 & -2 & 2 \\ 2 & -1 & 2 \\ 2 & -2 & 3 \end{array} \right) \; \; B = \; \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right) \; \; C = \; \left( \begin{array}{rrr} -1 & 2 & 2 \\ -2 & 1 & 2 \\ -2 & 2 & 3 \end{array} \right) $$

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I knew this but it didn't enter my mind when working on that other question. I wonder if it helps. –  mercio Feb 8 '13 at 20:53
    
@mercio, it does not give a two-parameter recipe, which may be unlikely as ciceksiz has suggested. But it does take you from any single point on $x^2 + y^2 - z^2 = k,$ which we need to find somehow, to any other such point. I would also expect a doubled or trebled ternary tree for any of these problems, as derived solutions may not have all positive entries. Worth trying to draw some trees, some for positive $k,$ some for negative, see how well it works out and what adjustments to the original picture may be required. Certainly "primitive" solutions will be the norm. –  Will Jagy Feb 8 '13 at 21:06

I think the answer to your question is no. There may be infinitely many solutions to the diophantine equations you have stated, however you cannot classify them as in the $x^2+y^2=z^2$ case. What I mean is that in the equations $x^2+y^2=z^2+n$, if you have a solution $(x_1,y_1,z_1)$, you cannot generalize this solution as to $(kx_1,ky_1,kz_1)$, since you have the factor n. For instance let us say we have the equation $x^2+y^2=z^2+12$. (5,6,7) is a solution of this equation, however when you plugin the values (10,12,14) the quation does not hold. So what I am saying is that you cannot find infinitely many solutions to the diaphontine equations $x^2+y^2=z^2+n$ with the same method you have applied in the phytogaros equation. Nevertheless, you may find infinitely many solutions using other parametrizations. Hope this helps you!

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For the special case when the number of is a square and is given us. That is, in the equation:

$X^2+Y^2=Z^2+q^2$

where the number of $q$ - given us.

Then the solutions of the equation can be written ospolzovavshis Pell: $p^2-2k(k-1)s^2=\pm{q}$

$k$ - given us can be anything.

$X=p^2+2(k-1)ps+2k(k-1)s^2$

$Y=p^2+2kps+2k(k-1)s^2$

$Z=p^2+2(2k-1)ps+2k(k-1)s^2$

If we use the solutions of Pell's equation: $p^2-2s^2=\pm1$ Solutions have the form:

$X=sp^2+2qps\pm(qp^2+2(p+q)s^2)$

$Y=2qps+2s^3\pm(qp^2+2(p+q)s^2)$

$Z=sp^2+4qps+2s^2\pm(qp^2+2(p+q)s^2)$

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