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Are there any solutions for $a, b, c$ such that:

$$a, b, c \in \Bbb N_1$$ $$\sqrt{a^2+(b+c)^2} \in \Bbb N_1$$ $$\sqrt{b^2+(a+c)^2} \in \Bbb N_1$$ $$\sqrt{c^2+(a+b)^2} \in \Bbb N_1$$

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    $\begingroup$ Is there a reason you would expect such examples exist? $\endgroup$ – Michael Burr Dec 21 '15 at 21:42
  • $\begingroup$ @Rasmus What is the purpose of this problem? $\endgroup$ – user266519 Dec 21 '15 at 21:44
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    $\begingroup$ No, I don't know if they exist. $\endgroup$ – Rasmus Dec 21 '15 at 21:45
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    $\begingroup$ I'm making math problems. $\endgroup$ – Rasmus Dec 21 '15 at 21:48
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    $\begingroup$ I know one thing: if $(a,b,c)$ is a primitive solution, then exactly one of $a$, $b$, and $c$ is odd, and the even numbers among them are divisible by $4$. $\endgroup$ – Batominovski Dec 21 '15 at 22:05
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$(108,357,368),(216,714,736)$ and its permutations are the unique solutions of your equation in positive integers below $1000$ (found using a computer search). Of course the latter triple is just a constant multiple of the former one. By multiplying by a constant we can find infinite number of solutions, but so far I have found only one primitive solution (i.e. with coprime numbers).

Edit: I have extended the range and I found one more primitive solution: $(564,748,1425)$. Right now I have looked at all triples with $a\leq b\leq c,a\leq 1000,b\leq 1500,c\leq 2000$.

Edit 2: Following Batominovski's suggestion on how to speed up calculation, I have ran a search with range $a\leq 3000$ odd, $b\leq c,b\leq 3000,c\leq 4000$ divisible by $4$ and here is the list of solutions in that range, in order of apprearance:

$$(357,108,368)\\ (975,348,2380)\\ (1071,324,1104)\\ (1425,564,748)\\ (1785,540,1840)\\ (2499,756,2576)$$

Here is the code I have used (quite ugly, I know):

#include<Windows.h>
#include<conio.h>
#include<iostream>
#include<Windows.h>
#include<math.h>

using namespace std;

long int main(){
    for(double a=1;a<3000;a+=2){
    for(double b=4;b<3000;b+=4){
    for(double c=b;c<4000;c+=4){
        if(floor(sqrt(a*a+(b+c)*(b+c)))==sqrt(a*a+(b+c)*(b+c))&&floor(sqrt(b*b+(a+c)*(a+c)))==sqrt(b*b+(a+c)*(a+c))&&floor(sqrt(c*c+(b+a)*(b+a)))==sqrt(c*c+(b+a)*(b+a))){
            cout<<a<<"|"<<b<<"|"<<c<<endl;
        }
    }
    }
    }
    cout<<"done";
    return getch();
}
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  • $\begingroup$ I think there is no other way do find these answers except a computer search, so +1. $\endgroup$ – SalmonKiller Dec 21 '15 at 22:00
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    $\begingroup$ Thank you for the answer! What is computer search and what kind of program did you use. @Wojowu $\endgroup$ – Rasmus Dec 21 '15 at 22:05
  • $\begingroup$ This problem can be asked in the form of three integer length lines that arranged in any permutation form a right triangle, which sounds counter-intuitive but the fact there is an answer is pretty impressive. Anyhow I wonder if there is a geometric approach that gives more insight. $\endgroup$ – Red Dec 21 '15 at 22:10
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    $\begingroup$ Checking all values under $10000$ yields the following four primitive triples, sorted in order of smallest element: $(108, 357, 368), (348, 975, 2380), (564, 748, 1425), (624, 4551, 6256)$. $\endgroup$ – Brian Tung Dec 21 '15 at 22:53
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    $\begingroup$ @Wojowu: One can also use elliptic curves to find new integer solutions. $\endgroup$ – Tito Piezas III Dec 22 '15 at 16:29
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Using an elliptic curve, it can be shown that the system,

$$a^2+(b+c)^2 = x_1^2\\b^2+(a+c)^2 = x_2^2\\c^2+(a+b)^2 = x_3^2\tag1$$

has an infinite number of integer solutions with $\gcd(a,b,c)=1$.

Method:

Let $a = m^2-n^2,\;b=2mn-c$ and $(1)$ becomes,

$$(m^2+n^2)^2 = x_1^2\\2c^2+2c(m^2-2mn-n^2)+(m^2+n^2)^2 = x_2^2\\2c^2-2c(m^2+2mn-n^2)+(m^2+2mn-n^2)^2 = x_3^2\tag2$$

Thus, the problem is reduced to a pair of quadratic polynomials in $c$ that is to be made a square. If there is a rational point $c$, then the pair is birationally equivalent to an elliptic curve and in general there should be an infinite more rational points.

For example, using the smallest solution $a,b,c = 357,\, 108,\, 368$ we get $m,n = 2,\, 1$, hence,

$$2 c^2-2c+25 = x_2^2\\ 2 c^2-14c+49 = x_3^2\tag3$$

A solution, of course, is $c=\frac{3\times108}{357}$. Using the tangent method, another one is,

$$c = \frac{2859837252}{16433685001}$$

though there may be smaller ones. Clearing denominators, we get a new solution in positive integers to $(1)$ as,

$$a = 49301055003\\ b = 2859837252\\c= 62874902752$$

More directly,

$$a = 357(2-x^2)\\ b = 2(1+54x)(11-x)\\c = 2(31+8x)(15-23x)$$

and $x$ satisfies,

$$F(x):=2703220 - 3847384 x + 424640 x^2 + 1463524 x^3 + 537289 x^4 = y^2\tag4$$

Since $(4)$ has a known rational point, then it is birationally equivalent to an elliptic curve. The point $x = 7/13$ yields the smallest $a,b,c$ (after clearing denominators), while $x = 82711/6095$ gives the new one.

Not all $x$ will yield positive $a,b,c$. But since $F(x)=y^2$ has an infinite number of rational solutions, with some hand-waving we may assume there is a small infinite subset that yields $a,b,c$ that are all positive.

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Here is a complete parametrization of all rational solutions to $\sqrt{a^2+(b+c)^2}\in\mathbb{Q}$, $\sqrt{b^2+(c+a)^2}\in\mathbb{Q}$, and $\sqrt{c^2+(a+b)^2}\in\mathbb{Q}$, where $a,b,c\in\mathbb{Q}$. If $p,q,r\in\mathbb{Q}_{\geq 0}$ be such that $$\frac{2p}{1+2p-p^2}+\frac{2q}{1+2q-q^2}+\frac{2r}{1+2r-r^2}=1\,,\tag{*}$$ then $(a,b,c)=\left(\frac{2p}{1+2p-p^2}x,\frac{2q}{1+2q-q^2}x,\frac{2r}{1+2r-r^2}x\right)$ for some $x\in\mathbb{Q}$ (namely, $x=a+b+c$). All rational solutions $(a,b,c)$ are of this form. There exists a positive integer solution $(a,b,c)$ associated to $(p,q,r)$ iff $0< p,q,r<1+\sqrt{2}$.

Frankly, I don't know if solving (*) is any easier than using the method mentioned by Tito Piezas III, but at least, there is one equation to be solved now, and with only $3$ rational variables. (However, if you try to write $p=\frac{m_1}{n_1}$, $q=\frac{m_2}{n_2}$, $r=\frac{m_3}{n_3}$, where $m_i,n_i\in\mathbb{Z}$ for $i=1,2,3$, then you will end up with $6$ variables, but the method mentioned by Tito Piezas III can reduce the number of variables to $5$.) There may be an algebraic-geometry/algebraic-number-theory method to solve (*), but I'm not so knowledgeable in these fields. Here is an example: $(a,b,c)=(108,357,368)$ is given by $(p,q,r,x)=\left(\frac{2}{27},\frac{1}{3},\frac{8}{23},833\right)$, where $\frac{2p}{1+2p-p^2}=\frac{108}{833}$, $\frac{2q}{1+2q-q^2}=\frac{3}{7}=\frac{357}{833}$, and $\frac{2r}{1+2r-r^2}=\frac{368}{833}$.

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  • $\begingroup$ I simplified my answer and gave a new primitive solution $a,b,c$ in positive integers. $\endgroup$ – Tito Piezas III Dec 22 '15 at 16:27
  • $\begingroup$ @TitoPiezasIII Do you know how to solve equation (*) or, at least, find an infinite family of $(p,q,r)$ that satisfy (*)? $\endgroup$ – Batominovski Dec 22 '15 at 17:22
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    $\begingroup$ Well, yes, there is an infinite family, $$p,q,r = \frac{-1 + m}{1 + m}, \, \frac{m (1 + 3 m^2 - 4 m^3 + 3 m^4 + 4 m^5 + m^6)}{1 - 4 m + 3 m^2 + 4 m^3 + 3 m^4 + m^6}, \, \frac{(1 - m) (1 + m) (1 + m^2)^2}{2 m (1 - 2 m + 2 m^2 + 2 m^3 + m^4)}$$ May I know what you need it for? $\endgroup$ – Tito Piezas III Dec 24 '15 at 3:48
  • $\begingroup$ @TitoPiezasIII, I was hoping to get an infinite family of $(p,q,r)$ with $p,q,r>0$ in order to create a parametrized family $(a,b,c)$ of positive integers that satisfy the conditions. Triples in your parametrized family $(p,q,r)$ usually contain a negative entry, though. It is great, nonetheless, as it gives an infinite parametrized family of integer solutions $(a,b,c)$ to the required conditions. Would you please post how you obtained this family of triples? $\endgroup$ – Batominovski Dec 24 '15 at 9:38
  • $\begingroup$ This comment box is too small. Ask it as a separate question, and I'll share the algebraic trick. :) $\endgroup$ – Tito Piezas III Dec 24 '15 at 13:43

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