Supposing that solution exists, I've found that $x$, $y$, $(x+y)$, $(x-y)$ should be pairwise coprime numbers, and hence $x$ and $y$ are coprime of different parity. And after that I can not make any step further.
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Not sure if this is what you are looking for, but $(\pm a,\pm a,0)$ are certainly solutions. :D – yo' Dec 07 '17 at 23:07
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No, I need every cofactor of the left part to be nonzero ) – Vladislav Romanovskiy Dec 07 '17 at 23:12
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2I you had the simpler $xy = 10 z^2,$ you would still be able to assume $x,y$ coprime, and then have several cases: $x = 10 s^2, y = t^2.$ Next $x = 5 s^2, y = 2 t^2,$ and so on. Your version is worse. The point is if pairwise coprime numbers multiply to become a square, each is a square, once they are all known to be positive – Will Jagy Dec 07 '17 at 23:25
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1Oh: computer run suggests impossibility. – Will Jagy Dec 07 '17 at 23:26
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4If $(x,y,z)$ is a solution, then so is $(x+y,x-y,2z)$. – 2'5 9'2 Dec 08 '17 at 00:14
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Following my last comment, the next iteration is that $(2x,2y,4z)$ is a solution. So that calls into question the claim that $x$ and $y$ are of different parity. Or, if you really did establish that, then here is your contradiction. – 2'5 9'2 Dec 08 '17 at 00:17
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1Or maybe you are saying that you may assume $x$ and $y$ are coprime of different parity because you can reduce other cases to that case? – 2'5 9'2 Dec 08 '17 at 00:23
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@alex could you please look at the revised problems I computed? Interesting the number of solutions I could find (in a reasonable time on the computer) goes up with prime factor number of the multiplier. – Will Jagy Dec 08 '17 at 01:11
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Idea: if we fix a value for either $x$ or $y$ and consider this as an equation the remaining two variables (if the fixed value is non-zero), so this question is asking for the integral points on a family of elliptic curves. Not sure if this viewpoint helps. – Lukas Heger Dec 08 '17 at 02:07
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@WillJagy You can see my comment in your data, in that the solution sets pair up. Like for 2730: $(10,3,z)$ with $(13,7,z)$ and $(13,8,z)$ with $(21,5,z)$. – 2'5 9'2 Dec 08 '17 at 02:10
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1@alex, good. I did a separate experiment with multiplier $1.$ I think there may be a complete proof available, as requiring the $x,y,x+y$ are squares is very strong, Pythagorean triple. – Will Jagy Dec 08 '17 at 02:39
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@Alex, yes, I mean I may assume $x$ and $y$ are coprime of different parity because other cases can be reduced to that. – Vladislav Romanovskiy Dec 08 '17 at 07:11
3 Answers
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The original question is equivalent to asking whether $10$ is a congruent number - which it isn't.
With a general multiplier $N$, we have \begin{equation*} xy(x+y)(x-y)=Nz^2 \end{equation*}
Define $t=x/y$ and $w=z/y^2$ giving \begin{equation*} t(t+1)(t-1)=Nw^2 \end{equation*} and then define $s=Nt$ and $r=N^2w$ leading to \begin{equation*} r^2=s^3-N^2s \end{equation*} which is the elliptic curve for the congruent number problem.
Non-zero solutions only come from curves with rank greater than zero. This includes $N=5,6,7$ which give curves of rank $1$. $N=2730$ gives a curve of rank $2$ using Denis Simon's ellrank code.
Allan MacLeod
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Sort of interesting. I replaced the $10$ with $2730$ in order to get at least one solution, it suddenly gave many:
Thu Dec 7 15:53:49 PST 2017
10 3 2730 = 2 3 5 7 13
13 7 10920 = 2^3 3 5 7 13
13 8 10920 = 2^3 3 5 7 13
14 1 2730 = 2 3 5 7 13
15 13 10920 = 2^3 3 5 7 13
21 5 43680 = 2^5 3 5 7 13
1573 128 494892478080 = 2^7 3^5 5 7 11^2 13 17^2
1694 1681 124939064250 = 2 3^3 5^3 7 11^2 13 41^2
1701 1445 1979569912320 = 2^9 3^5 5 7 11^2 13 17^2
3375 13 499756257000 = 2^3 3^3 5^3 7 11^2 13 41^2
3610 3267 27819510288570 = 2 3^3 5 7^3 11^2 13 19^2 23^2
6877 343 111278041154280 = 2^3 3^3 5 7^3 11^2 13 19^2 23^2
Thu Dec 7 15:55:10 PST 2017
===================================================
Multiplier set to $6$
Thu Dec 7 16:47:57 PST 2017
2 1 6 = 2 3
3 1 24 = 2^3 3
25 24 29400 = 2^3 3 5^2 7^2
49 1 117600 = 2^5 3 5^2 7^2
2738 529 10452831898806 = 2 3^3 11^2 23^2 37^2 47^2
3267 2209 41811327595224 = 2^3 3^3 11^2 23^2 37^2 47^2
Thu Dec 7 16:50:07 PST 2017
============================================================
Multiplier set to $7,$ and checked for $0<y < x < 11000$
Thu Dec 7 16:56:20 PST 2017
16 9 25200 = 2^4 3^2 5^2 7
25 7 100800 = 2^6 3^2 5^2 7
Thu Dec 7 16:58:08 PST 2017
================================================================
Multiplier set to $14$
Thu Dec 7 17:15:36 PST 2017
8 1 504 = 2^3 3^2 7
9 7 2016 = 2^5 3^2 7
4225 2016 117426776594400 = 2^5 3^2 5^2 7 13^2 47^2 79^2
6241 2209 469707106377600 = 2^7 3^2 5^2 7 13^2 47^2 79^2
=========================================
Will Jagy
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The reference to congruent numbers settles the original question and similar questions with $10$ replaced with $N$. There is an even more general result relating to division polynomials and discrete analogs of Weierstrass sigma and Jacobi theta functions. This is given in my work at Weierstrass Elliptic Function Polynomials. In brief, I construct four sequences $\, (w_n,x_n,y_n,z_n) \,$ such that $\, (x_n/x_0)^2 - (y_n/y_0)^2 = (x_1^2-y_1^2) (w_n/w_1)^2, \,$ $\, (z_n/z_0)^2 - (y_n/y_0)^2 = (z_1^2-y_1^2) (w_n/w_1)^2, \,$ $\, (z_n/z_0)^2 - (z_n/z_0)^2 = (z_1^2-x_1^2) (w_n/w_1)^2 \,$ for all integer $\, n.$ They satisfy recursions such as $\, w_{n+1}w_{n-1}x_0^2 = w_n^2 x_1^2 - x_n^2 w_1^2, \,$ and $\, x_{n+1}x_{n-1}x_0^2 = x_n^2 x_1^2 - (x_1^2-y_1^2)(x_1^2-z_1^2)(w_n/w_1)^2 x_0^4. $
For an numerical example we let $$ x_0 \!=\! y_0 \!=\! z_0 \!=\! w_1 \!=\! y_1 \!=\! 1 \quad \text{and} \quad x_1 \!=\! \sqrt{2},\, z_1 \!=\! \sqrt{3},\, x_1^2-y_1^2 \!=\! 1,\, z_1^2-y_1^2 \!=\! 2,\, z_1^2-x_1^2 \!=\! 1. $$ In this particular example we have $$ x_n^2-y_n^2 = w_n^2, \,\, z_n^2-y_n^2 = 2w_n^2, \,\, z_n^2-x_n^2 = w_n^2, \,\, x_n^2-w_n^2 = y_n^2, \,\, x_n^2+w_n^2 = z_n^2. $$ This implies $\, x_n^2 w_n^2 (x_n^2+w_n^2)(x_n^2-w_n^2) = (w_nx_ny_nz_n)^2 = w_{2n}^2 /4. \,$ Because $\, w_n \,$ is a divisibility sequence, we have $\, w_{2n} = w_2t_n \,$ where $\, w_2 = 2\sqrt{6} \,$ and $\, t_n \,$ is an integer sequence. Thus, $\, w_{2n}^2/4 \!=\! 6\,t_n^2. $ This gives a sequence of solutions to $\, x^2 w^2(x^2+w^2)(x^2-w^2) = 6\,t^2. \,$ The four sequences $\, w_n^2, x_n^2, y_n^2, z_n^2 \,$ are integer sequences.
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