0
$\begingroup$

How do I show that if $$Dm^2 - n^2D^2$$ is a perfect square for some integers $m$ and $n$ ($n \neq 0$), $D$ is the sum of two (non-zero) perfect squares? I tried solving for $D$ but that only gives me $$D = \frac{m^2}{2n^2} \pm \frac{\sqrt{m^4 - 4n^2 k^2}}{2n^2}$$ for integers $m$, $n$, and $k$, which doesn't seem easier.

EDIT: $D$ itself should not be a perfect square.

$\endgroup$
10
  • $\begingroup$ But, he didn't say if and only if. It might be a sufficient condition but not necessary. @GyuminRoh $\endgroup$ – MathIsNice1729 Dec 3 '15 at 7:04
  • $\begingroup$ If $Dm^2 - n^2D^2$ is a perfect square for any integers $m,n$, then in particular, letting $m = 1, n = 0$, we have that $D$ is a perfect square. But then $D = D + 0^2$ is a sum of two perfect squares. $\endgroup$ – Ethan Alwaise Dec 3 '15 at 7:05
  • $\begingroup$ Oops misread the question sorry. I think the above solution works. $\endgroup$ – Gyumin Roh Dec 3 '15 at 7:07
  • $\begingroup$ But, is the question really saying this? I mean, can we just let $m$ and $n$ be some fixed numbers? I think we should go for a general approach. I believe it'd be easier to prove contrapositively. @EthanAlwaise $\endgroup$ – MathIsNice1729 Dec 3 '15 at 7:09
  • $\begingroup$ @EthanAlwaise I edited the question to make it more precise. n should not be 0. $\endgroup$ – arbitrary username Dec 3 '15 at 7:09
3
$\begingroup$

If $Dm^2-D^2n^2= a^2$, then $Dm^2$ is a sum of two squares. Now an integer is a sum of two squares if and only if all primes $\equiv 3 \mod 4$ in its factorization occur with even multiplicities. The presence of the extra square $m^2$ doesn't affect this condition, so $D$ is a sum of two squares, also.

$\endgroup$
8
  • 1
    $\begingroup$ Also the hypothesis $n \neq 0$ is unnecessary. If the statement is true for some $m$ when $n = 0$ then $Dm^2$ is a perfect square. But then $D$ must be a perfect square and so $D = D + 0^2$ is a sum of two squares. $\endgroup$ – Ethan Alwaise Dec 3 '15 at 7:32
  • $\begingroup$ How do you prove that an integer is the sum of two squares if and only if all primes $\equiv 3$ mod $4$ in its factorization occur with even multiplicities? $\endgroup$ – arbitrary username Dec 3 '15 at 7:35
  • $\begingroup$ It's not obvious, you can get it as a by-product of the analysis of Gaussian primes: en.wikipedia.org/wiki/Gaussian_integer $\endgroup$ – user138530 Dec 3 '15 at 7:38
  • $\begingroup$ And see here: en.wikipedia.org/wiki/… $\endgroup$ – user138530 Dec 3 '15 at 7:41
  • 1
    $\begingroup$ @GerryMyerson: I'm not claiming I need it, I'm just saying it's a quick way to finish it off. $\endgroup$ – user138530 Dec 3 '15 at 9:10
0
$\begingroup$

The proposition is false. Suppose every prime divisor (if any) of $D$ is congruent to $3$ modulo $4$ and that $D$ is a square. Then $D$ is not the sum of two non-zero squares. Suppose $(p,q,r)$ is any Pythagorean triplet with $p^2=q^2+r^2$. Let $D=E^2$. We have $D(Ep)^2-q^2D^2=(rD)^2.$ Examples:(I)D=1,m=5,n=4. (II).D=9,m=15,n=4.

$\endgroup$
4
  • 1
    $\begingroup$ But the problem says $D$ is not to be a square. $\endgroup$ – Gerry Myerson Dec 3 '15 at 9:27
  • $\begingroup$ The edit demanding that D is a non-square wasn't there before. $\endgroup$ – DanielWainfleet Dec 3 '15 at 9:31
  • 1
    $\begingroup$ The last edit on the question was 2 hours ago. Your answer went up 52 minutes ago. $\endgroup$ – Gerry Myerson Dec 3 '15 at 9:38
  • $\begingroup$ SO WHAT .I thought about the question for along while and put up an answer $\endgroup$ – DanielWainfleet Dec 3 '15 at 14:07
0
$\begingroup$

Let's take the equation.

$$dx^2-d^2y^2=z^2$$

If we represent the coefficient as a sum of squares. $d=a^2+b^2$

The solution can be written as.

$$x=d(p^2+s^2)$$

$$y=ap^2+2bps-as^2$$

$$z=d(bs^2+2aps-bp^2)$$

$\endgroup$

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.