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show that : there exist infinitely many values of postive rational numbers $(a,b,c)$ such $$\begin{cases} a+b+c=3.78\\ abc=2 \end{cases}$$

ie: show this equation $$x^3-3.78x^2+kx-2=0$$ has infinitely many postive rational numbers roots$(a,b,c)$ such example $a=b=1.25,c=1.28$.because $$a+b+c=1.25+1.25+1.28=3.78,abc=1.25\cdot 1.25\cdot 1.28=2$$

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  • $\begingroup$ A system of two equations and three unknowns has an infinite number of solutions... $\endgroup$ – Martigan Jun 25 '15 at 12:26
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    $\begingroup$ but some system of two equation can't infinite rational numbers, for example $a+b+c=3,abc=1$ this system has only postive solution $a=b=c=1$ $\endgroup$ – user225250 Jun 25 '15 at 12:29
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    $\begingroup$ @Martigan I'd be interested to see even one solution of the system $$ a+b+c = 1 \\ a+1=a+2$$ $\endgroup$ – Surb Jun 25 '15 at 12:34
  • $\begingroup$ other words,some system of two equation there is no need to have an infinite number of groups that need to look at the right side of the specific number,my problem the number $3.78$ and $2$ is special,But I can't How to use the key is special $\endgroup$ – user225250 Jun 25 '15 at 12:35
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There is a result which gives a condition concerning the existence of infinitely rational solutions of the equations $x+y+z=r$ and $xyz=s$ (the variables are here called $x,y,z$ and not $a,b,c$). It is Theorem $1.2$ here:

Theorem 1.2: Let $a,b,c$ be pairwise distinct rational numbers such that, for every permutation $(A,B,C)$ of $(a,b,c)$, we have $$ A(B-C)^3\neq B(C-A)^3 $$ and $$ AB^2+BC^2+CA^2\neq 3ABC. $$ Then there are infinitely many triples $(x,y,z)$ of rational numbers such that $$ x + y + z = a + b + c, \quad xyz = abc. $$

The result relies on exhibiting families of positive-rank elliptic curves. The possible triples are further classified in Proposition $4.2$.

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