Do there exist postive integers $a,b,c,x,y,p,q$ such $(a,b,c)$, $(x,y,a)$, $(p,q,b)$ are all Pythagorean triples? That is, does the system $$\begin{cases} a^2+b^2=c^2\\ x^2+y^2=a^2\\ p^2+q^2=b^2 \end{cases}$$ have a postive integer solution?
Is it possible that $(a,b,c)$, $(x,y,a)$, $(p,q,b)$ are Pythagorean triples simultaneously? [closed]
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2$\begingroup$ There can't be three primtive triples, but there can be two with an additional triple satisfying the relations. $\endgroup$– The Masked AvengerCommented Jun 10, 2015 at 3:30
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4$\begingroup$ Using the parametrization of Pythagorean triples as $(u^2-v^2,2uv,u^2+v^2)$ it's pretty quick to generate solutions by hand such as $a = 25$, $b = 312$; i'll leave it to you to fill in the other variables from there. $\endgroup$– ARupinskiCommented Jun 10, 2015 at 3:44
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2 Answers
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Yes, for example $$ (a,b,c,x,y,p,q) = (145,10512,10513,143,24,7920,6912).$$
P.S. I found this example by looking for $(u,v)$ such that $u,v,u+v,u-v$ are all sums of two squares (cf. ARupinski's comment). I took $u=73$ and $v=72$.
answered Jun 10, 2015 at 3:42
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The system of equations.
$$\left\{\begin{aligned}&a^2+b^2=c^2\\&x^2+y^2=a^2\\&f^2+q^2=b^2\end{aligned}\right.$$
Equivalent to the need to solve the following system of equations.
$$\left\{\begin{aligned}&a=2ps=z^2+t^2\\&b=p^2-s^2=j^2+v^2\\&c=p^2+s^2\\&x=2zt\\&y=z^2-t^2\\&f=2jv\\&q=j^2-v^2\end{aligned}\right.$$
To ease calculations we change.
$$B=k^2+2n^2-r^2$$
$$A=k^2+2n^2+r^2-4nr$$
$$W=2k(r-2n)$$
$$Q=2n^2+r^2-k^2-2rn$$
Then we need the number to obtain Pythagorean triples can be found by the formula.
$$p=B^2+A^2$$
$$s=2Q^2$$
$$z=2BQ$$
$$t=2AQ$$
$$j=B^2+A^2-2Q^2$$
$$v=2WQ$$
At any stage of the computation can be divided into common divisor.
