1

My approach so far:

Let $x^2+y^2+xy=n^2,$ where $n\in\mathbb Z$.

$\implies (x+y) ^2-xy=n^2$

$\implies (x+y) ^2-n^2=xy$

$\implies (x+y+n) (x+y-n) =xy$

Only one case is possible:

When $x+y+n=xy$ and $x+y-n=1$. On adding, we get: $2(x+y) =xy+1$. Since LHS is an even number so is RHS. That's why, $xy$ must be odd$\implies x$ and $y$ both are odds. So I checked for few odd numbers and found out that for $x=3$ and $y=5,\;x^2+y^2+xy$ becomes $49$ which is a perfect square.. I didn't dare to check for more odds as there could be many...

Recently I found out that for $x=5$ and $y=16$(an even number), $x^2+y^2+xy=361=19^2.$ (Surprising!!)

So now I can say, I am stuck very badly.. All of my observations miserably failed...

Please suggest something!

Dhrubajyoti Bhattacharjee
  • 1,265
  • Why only one case is possible ? – Student May 23 '20 at 22:12
  • Your reasoning that "only one case is possible" is not true. I mean, you provided a counterexample yourself: for $x = 5, y = 16, n = 19$, we have $40 = x+y+n \neq xy = 80$, and $x + y - n = 2 \neq 1$. – twosigma May 23 '20 at 22:22
  • Consider $x+y+n=1$ and $x+y-n=xy$. But certainly $x+y+n\ne 1$ as $x, y\in\mathbb N$ and $n\in\mathbb Z$. Now consider $x+y+n=x$ and $x+y-n=y\implies y=-n$ and $x=n$ but as $x, y\in\mathbb N$ so this case is also not possible because here either $x$ or $y$ is negative for $n\in\mathbb Z$. A similar argument holds for the case, $x+y+n=y$ and $x+y-n=x$. Thus only one case(mentioned it already) is possible. – Dhrubajyoti Bhattacharjee May 23 '20 at 22:28
  • 2
    @DhrubajyotiBhattacharjee You argument only really applies of $x$ and $y$ must be primes, but didn't specify this, and your example of $x = 5$ and $y = 16$ indicates this is not an unstated requirement. With your specific example, you also have $n = 19$, with $5 + 16 + 19 = 40 = 5(8)$ and $5 + 16 - 19 = 2$, so $40(2) = 80 = 5(16)$, with the factors of $y$ being split between those $2$ factors of $x + y - n$ and $x + y + n$. – John Omielan May 23 '20 at 22:33
  • https://en.wikipedia.org/wiki/Integer_triangle#Integer_triangles_with_a_120.C2.B0_angle – Will Jagy May 23 '20 at 22:39
  • Got it @John Omielan.. Thanks. Will try to work on it. – Dhrubajyoti Bhattacharjee May 23 '20 at 22:45
  • @Anurag A. Yes.. But at the moment I'm not able to understand the answers completely. Gotta read for it. Thanks.. – Dhrubajyoti Bhattacharjee May 23 '20 at 22:47
  • 1
    Dhrub, All are from coprime $u,v$ and then $x = u^2 - v^2, ; ;$ $ y = 2uv + v^2 ; , ; ; $ and your $n = u^2 + uv + v^2.$ Or negate all three. The approach goes back to an 1897 book by Fricke and Klein. Confirm that my $x^2 + xy + y^2 = n^2 ; ; ; ; !$ – Will Jagy May 23 '20 at 22:56
  • @Will Jagy, yes sir, I solved it. Just divided both sides of $x^2+y^2+xy=n^2$ by $n^2$ to get an equation of the form $X^2+Y^2+XY=1$. Clearly $(1, 0) $ lies on this ellipse so solving it with the line $y=m(x-1) $ gives the other rational point viz., $X=(1-m^2) /(m^2+m+1) $ and $Y=(m^2+2m) /(m^2+m+1) $. Now letting $m=u/v$ yields the same general solutions that you have mentioned. And of course, $\gcd(u, v) =1.$ Thanks... – Dhrubajyoti Bhattacharjee May 23 '20 at 23:33

2 Answers2

1

$x^2+yx+y^2-n^2=0 \implies \triangle = 4n^2-3y^2= d^2$. At this point I propose you to look up a article by L.J. Mordell or Kneser in internet and they have a formula for all the finite solutions that are bounded above by the products of the coefficients $(4,-3,-1)$ of the Diophantine equation above.

DeepSea
  • 77,651
  • https://en.wikipedia.org/wiki/Integer_triangle#Integer_triangles_with_a_120.C2.B0_angle – Will Jagy May 23 '20 at 22:40
  • Sure.. I'll look for it. Thank you. – Dhrubajyoti Bhattacharjee May 23 '20 at 22:45
  • @Will Jagy: Kneser and Mordell in 1959 wrote the papers that solve this type of equation . The one you listed are for Pythagorean triples while this is not. – DeepSea May 23 '20 at 22:45
  • 1
    All are from coprime $u,v$ and then $x = u^2 - v^2, ; ;$ $ y = 2uv + v^2 ; , ; ; $ and your $n = u^2 + uv + v^2.$ Or negate all three. The approach goes back to an 1897 book by Fricke and Klein. Confirm that my $x^2 + xy + y^2 = n^2 ; ; ; ; !$ – Will Jagy May 23 '20 at 23:04
1

Firstly, we can ssume Multiplying by $4$, we get $(2x+y)^2+3y^2$ is a perfect square. Thus we want to find $a,b$ such that $a^2+3b^2$ is a perfect square with $gcd(a,b)=1$

What follows is a well-known theory of finding out rational points on conic sections.

Now this amounts to finding rational points on the ellipse $E: X^2+3Y^2=1$. We know $(1,0)$ is a rational point. If $P$ is another such point, then the slope of the line joining $P$ and $(1,0)$ is rational $m$. Say $L$ intersects $E$ at $(u,v)$. Then we get $$ \frac{v}{u-1}=m $$ and $$u^2+3v^2=1$$ This gives us $u^2+3m^2(u-1)^2=1$ Since $1$ ia anyway a root of the quadratic, the other root is $\frac{3m^2-1}{3m^2+1}$ and this gives $u=\frac{3m^2-1}{3m^2+1},v=\frac{-2m}{3m^2+1}$

Thus all rational points are parametrized by $\left (\frac{3m^2-1}{3m^2+1},\frac{-2m}{3m^2+1} \right ) \ ; \ m\in \mathbb Q$ or $\left (\frac{3m^2-1}{3m^2+1},\frac{2m}{3m^2+1} \right ) \ ; \ m\in \mathbb Q$

So we get if $a^2+3b^2=c^2$, then $\frac{a}{c}=\frac{3m^2-1}{3m^2+1}$ $\frac{b}{c}=\frac{2m}{3m^2+1}$ for some $m\in \mathbb Q$ writing everything in terms of integers we get $$a=\pm(3p^2-q^2),b=\pm 2pq, c= \pm (3p^2+q^2)$$

Thus we get $y=6kpq$ where $p,q,k>0$ and $2x+y=k|3p^2-q^2|$

If $p,q$ have opposite parity, this forces $k$ to be even and we get $k=2l$

So we finally get $y=4lpq$ and $x=l|3p^2-q^2|-2lpq$

If, on the other hand both $p,q$ are odd then we get $y=2kpq, x=\frac{k}{2}|3p^2-q^2|-kpq$

user6
  • 4,092