Find all natural numbers $x$ and $y$ such that $x^2+6xy+y^2$ is a square number.
For example, $(x,y)=(2,3)$ or $(x,y)=(3,10)$.
Obviously, we can consider $gcd(x,y)=1$.
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Let $s=x+y$ and $p=xy$. Then $s^2+4p=n^2$. But this is the $\Delta$ of $t^2-st-p=0$. When $\Delta=n^2$, the quadratic equation has two rational roots. At the same time, x and y are the $($integer$)$ roots of $u^2-su+p=0$. – Lucian Sep 20 '14 at 03:22
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The solution there. http://math.stackexchange.com/questions/816681/find-all-integers-satisfying-m2-n-12n-1n-2n-22/816685#816685 – individ Sep 20 '14 at 04:13
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Stereographic projection from $(-1,0),$ I get $$ x = m^2 - n^2, \; \; y = 2 mn + 6 n^2, $$ with $$ \gcd(m,n) = 1, $$ and either $m > n > 0$ or $n < 0$ and $|n| < m < 3 |n|. $ Also $m,n$ not both odd.
Put the inequalities together, we get $$ m > n > 0 \; \; \mbox{OR} \; \; \frac{-m}{3} >n > -m. $$ In the latter case we are originally in the third quadrant as the rational point on the hyperbola $x^2 + 6xy+y^2 = 1$ is $$ x= \frac{m^2 - n^2}{m^2 + 6 mn+n^2}, \; \; y= \frac{2mn+6n^2}{m^2 + 6 mn+n^2}, $$ while $-m/3 > n > -m$ implies $m^2 + 6 mn + n^2 < 0.$
Improvement: If $m,n$ both odd, take $$ x = \frac{m^2 - n^2}{4}, \; \; y = \frac{2 mn + 6 n^2}{4}, $$ final requirement here is that $m \neq n \pmod 4.$ Put another way: when both are odd, we require that $m+n$ be divisible by $4.$
The nonsense with the possible negative signs comes from the fact that we are not projecting onto an ellipse, we are projecting onto a hyperbola. So, the denominator may come up negative and solutions disappear. Need to allow $n$ negative. Let me know if I have missed any solutions; fairly easy to just splat a formula on the page, harder to figure out whether one has all solutions.
Note that, given some pair $(x,y)$ such that $x^2 + 6 xy+ y^2 = w^2$ for some $w,$ we get new pairs ( same $w$) with $$ (-y,x+6y), $$ $$ (-x-6y,6x+35y), $$ $$ (-6x-35y,35x+204y), $$ $$ (-35x-204y,204x+1189y) $$ and so on, which pushes along the relevant hyperbola in the second or fourth quadrant. As you will see, there is some repetition below. First ordered by $m,n$ then a short thing ordered by just $x,y.$
EDIT: this worked out very well. I just restricted to those solutions with $x>y$ and put them in order, well, backwards...Being in order, I was able to compare with a list of all solutions with $x \leq 306,$ we have a winner.
x y m n
306 19 35 1
300 13 37 -13
297 80 19 -8
290 171 39 -19
286 279 35 9
285 92 17 2
275 42 18 -7
273 232 17 4
270 119 37 -17
261 220 19 -10
260 189 33 7
255 38 16 1
253 12 17 -6
247 150 16 3
240 17 31 1
234 115 31 5
230 39 33 -13
225 112 17 -8
221 84 15 2
216 209 35 -19
210 11 31 -11
208 57 29 3
207 70 16 -7
200 153 33 -17
198 175 29 7
195 34 14 1
187 138 14 3
184 105 31 -15
182 15 27 1
176 105 27 5
171 10 14 -5
168 65 29 -13
165 76 13 2
161 144 15 -8
154 51 25 3
152 33 27 -11
143 30 12 1
136 9 25 -9
133 60 13 -6
132 13 23 1
126 95 23 5
119 30 12 -5
117 68 11 2
114 91 25 -13
105 8 11 -4
102 55 23 -11
99 26 10 1
90 11 19 1
85 84 11 -6
78 7 19 -7
77 60 9 2
70 39 17 3
65 24 9 -4
63 22 8 1
56 9 15 1
55 6 8 -3
52 45 17 -9
44 21 15 -7
40 33 13 3
36 5 13 -5
35 18 6 1
30 7 11 1
21 4 5 -2
15 14 4 1
12 5 7 1
10 3 7 -3
3 2 2 -1
x y m n
3 2 2 -1
3 10 2 1
5 12 3 -2
2 3 3 1
5 36 3 2
7 30 4 -3
15 14 4 1
7 78 4 3
9 56 5 -4
21 4 5 -2
21 44 5 2
4 21 5 3
9 136 5 4
11 90 6 -5
35 18 6 1
11 210 6 5
13 132 7 -6
33 40 7 -4
10 3 7 -3
12 5 7 1
45 52 7 2
33 152 7 4
6 55 7 5
13 300 7 6
15 182 8 -7
39 70 8 -5
55 6 8 -3
63 22 8 1
55 102 8 3
39 230 8 5
15 406 8 7
17 240 9 -8
14 15 9 -5
65 24 9 -4
77 60 9 2
65 168 9 4
8 105 9 7
17 528 9 8
19 306 10 -9
51 154 10 -7
99 26 10 1
91 114 10 3
51 434 10 7
19 666 10 9
x y m n
x y
1 0
3 2
10 3
12 5
15 14
21 4
30 7
35 18
36 5
40 33
44 21
52 45
55 6
56 9
63 22
65 24
70 39
77 60
78 7
85 84
90 11
99 26
x y
jagy@phobeusjunior:
I was curious about repetition, still with positive $x,y,$ to $x^2 + 6 xy+ y^2 = z^2.$ Plenty, and predictable, squarefree products of primes $p \equiv \pm 1 \pmod 8.$
x y m n z
65 24 9 -4 -119 7 . 17
90 11 19 1 119
99 26 10 1 161 7 . 23
136 9 25 -9 -161
143 30 12 1 217 7 . 31
102 55 23 -11 -217
351 14 20 -7 -391 17 . 23
176 105 27 5 391
2628 37 109 -37 -2737 7 . 17 . 23
1927 318 44 3 2737
1333 660 37 6 2737
1025 912 33 8 2737
81807 1030 304 -103 -84847 7 . 17 . 23 . 31
62625 8528 271 -104 -84847
59998 9735 491 33 84847
55040 12177 471 41 84847
49773 15052 247 -106 -84847
45140 17877 429 59 84847
39195 22018 226 -109 -84847
35032 25353 383 81 84847
x y m n z
Will Jagy
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1It might be worth emphasizing that this is intersecting the pencil of lines through $(-1,0)$ (i.e., $y=ax+a$) with the ellipse $x^2+6xy+y^2=1$, since rational points $\langle\frac sw,\frac tw\rangle$ on the latter correspond to natural numbers $s$, $t$, $w$ such that $s^2+6st+t^2=w^2$. – Steven Stadnicki Sep 19 '14 at 22:28
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4@StevenStadnicki, yes, but it is a hyperbola. I am still checking whether these points suffice, there is an unpleasant bit of trickery as the number to be squared has no fixed sign, different from Pythagorean triples for example. – Will Jagy Sep 19 '14 at 22:40
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For non-trivial cases, $xy\ne0$
Let $x^2+6xy+y^2=(x+ky)^2$ where $k$ is any integer
$\iff y(6x+y)=y(2kx+k^2y)\implies6x+y=2kx+k^2y\iff x(6-2k)=y(k^2-1)$
So, $\dfrac x{k^2-1}=\dfrac y{6-2k}=m$(say an integer)
As $x,y>0$ if $m<0,$ we need $6-2k<0\iff k>3\ \ \ \ (1)$ and $k^2-1<0\iff -1<k<1\ \ \ \ (2)$
There can be no $k$ satisfying both $(1),(2)$
Similarly if $m>0,$ we need $6-2k>0\iff k<3$ and $k^2-1>0\iff k>1$ or $k<-1$
$k<3,k<-1\implies k<-1$
or $k<3,k>1\implies1<k<3\implies k=2$
So, there are infinitely natural values of $x,y$
lab bhattacharjee
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