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Given real numbers $a, b, c$ such that $a^2= b^2+c^2$, there exists three sequences of natural numbers $a_n, b_n, c_n$ such that $a_n(a_n+1)= b_n(b_n+1)+c_n(c_n+1)$. The ratios $b_n/a_n$ and $ c_n/a_n$ converge to $b/a$ and $c/a$ respectively.

Can any one help how to get existence of such sequence and converges

GoodDeeds
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user90533
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  • Can any one help how to get existence of such sequence and converges – user90533 Jun 28 '16 at 10:46
  • Dunno if it helps, but completing the squares transforms the second equation into the form $-x^2+y^2+z^2=1$. Applying Gauss' theory of ternary quadratic forms, the solutions for this are parameterized by $(x,y,z) = (a^2+b^2+ac+bd, a^2-b^2+ac-bd, 2ab+ad+bc)$ and $(x,y,z) = (\frac{1}{2}(a^2+b^2-c^2-d^2), \frac{1}{2}(a^2-b^2-c^2+d^2), ab-cd)$, where $a,b,c,d$ are any integers such that $ad-bc=1$ that make the expressions integers. – Lance Sackless Jun 30 '16 at 13:59

1 Answers1

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Let $a_{n}=n$, $b_{n}=\frac{b}{a}n$, and \begin{align} c_{n}=\frac{1}{2}\left(\sqrt{\frac{4c^{2}}{a^{2}}n^{2}+4\left(1-\frac{b}{a}\right)n+1}-1\right) \end{align} which satisfies the given relations. We can easily check that $\lim_{n\to \infty} \frac{c_{n}}{a_{n}}=\frac{c}{a}$.

Seewoo Lee
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