11
$\begingroup$

For which n can $a^{2}+(a+n)^{2}=c^{2}$ be solved, where $a,b,c,n$ are positive integers? I have found solutions for $n=1,7,17,23,31,41,47,79,89$ and for multiples of $7,17,23$... Are there infinitely many prime $n$ for which it is solvable?

$\endgroup$
  • $\begingroup$ The title says "primitive", so $GCD(a,n,c)=1$ is assumed? $\endgroup$ – pharmine Oct 7 '11 at 12:35
  • 2
    $\begingroup$ @Angela: Very interesting question! $\endgroup$ – rubik Oct 7 '11 at 12:39
12
$\begingroup$

The general primitive solution to $x^2+y^2 = z^2$ is given by: $x=u^2-v^2$, $y=2uv$, $z=u^2+v^2$, with $u,v$ relatively prime and not both odd.

For $(a,a+n,z)$ to be a primitive triple, we'd have to have a $(u,v)$ such that: $|u^2 - v^2 - 2uv| = n$. We can rewrite that as: $(u-v)^2 - 2v^2 = \pm n$

So, setting $w = u-v$, we want to find $(w,v)$ which are relatively prime and $w$ is odd, with:

$$w^2-2v^2 = \pm n$$

This means that $n$ must be odd.

In fact, we can use unique factorization in $\mathbb{Z}[\sqrt{2}]$ to show that $n$ can be any product of primes of the form $8k\pm 1$. Since there are infinitely many primes of the form $8k\pm 1$, the answer to your question is, "yes."

(Oh, and once you find one solution $(w,v)$ for a particular $n$, you can find infinitely many solutions for that $n$.)

$\endgroup$
  • $\begingroup$ For example, $-71 = w^2 - 2v^2$ has solution $(w,v)=(1,6)$. So $u=7$, $x=u^2-v^2 = 13$, $y=2uv=84$, and $z=u^2+v^2=85$, so $n=71$ has a solution with $a=13$. $\endgroup$ – Thomas Andrews Oct 7 '11 at 19:01
  • $\begingroup$ Explicitly, if $p^2 + (p+n)^2 = r^2$, then subsequent ones can be found as $q^2 + (q+n)^2 = (p+q+r+n)^2$, where $q = 3p+2r+n$. $\endgroup$ – Tito Piezas III Nov 25 '14 at 23:40
1
$\begingroup$

$2a^{2}+2na+n^{2}=c^{2}$ --> $a=-\frac{-n+\sqrt{2c^{2}-n^{2}}}{2}$ --> there are solutions iff $x^{2}+n^{2}=2c^{2}$ has solutions --> find the set of the squares of all integers 0 in the set such that $y=2x-n$ then there is a primitive pythagorean triple with a difference of n between legs, and also for any multiple An if n>1 since if $k^{2}\equiv x(\mod{n})$ then $(Ak)^{2} \equiv Ax(\mod{An})$ --> $Ay=2Ax-An$.

$\endgroup$
-1
$\begingroup$

If you solve expression for $n$ you get

$n=\sqrt{c^2-a^2}-a$, let's denote $b=\sqrt{c^2-a^2}$,so we have that $n=b-a$

Now,take look at picture bellow.Note that $AD=a$,and $BD=b-a=n$

If you change value of $b$ and keep $a$ to be constant you will get a infinite number of right triangles,and therefore infinite number of values of $n=b-a$,so answer is yes, there are infinitely many primes $n$ for which equation is solvable.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Except that you don't know for which integer values of $b$ yield integer values of $c$. $\endgroup$ – Thomas Andrews Oct 7 '11 at 14:57
  • $\begingroup$ @Thomas,That's true,I have answered only on second part of the question...however, one can find one (a,b,n) triple and then for each different triple choose another b such that n becomes prime number $\endgroup$ – Peđa Terzić Oct 7 '11 at 15:18
  • $\begingroup$ So, what does your argument say when $n=3$? $\endgroup$ – Thomas Andrews Oct 7 '11 at 15:33
  • $\begingroup$ @Thomas,I didn't say that this reasoning is correct for each prime number.I just pointed that there is infinitely many distinct (a,b,n) triples such that n is prime... $\endgroup$ – Peđa Terzić Oct 7 '11 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.