Solve in integers $$x+\frac1{y+\frac1z}=\frac{10}7$$
My work:
1) $$\frac{xyz+x+z}{yz+1}=\frac{10}7$$
2) Solutions: $x=1, y=2, z=3$ and $x=2, y=-2, z=4$
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Your work looks correct, but I just need to verify it. – Toby Mak Dec 05 '19 at 08:47
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2For positive integers, of course $$\frac{10}{7}=1+\frac{3}{7}=1+\frac{1}{\frac{7}{3}}=1+\frac{1}{2+\frac{1}{3}}$$ and you may use the uniqueness of continued fractions, canonical form. – rtybase Dec 05 '19 at 08:48
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When $x=1$:
$$\frac{yz+1+z}{yz+1} = \frac{10}{7}$$ $$7yz + 7 + 7z = 10yz+10$$ $$3yz-7z+3=0$$ $$z=-\frac{3}{3y-7}$$
and setting $3y-7$ equal to $-3,-1,1$, and $3$ gives the only solution $(1, 2, 3)$.
And when $x=2$: $$\frac{2yz+2+z}{yz+1} = \frac{10}{7}$$ $$14yz + 14 + 7z = 10yz+10$$ $$4yz+7z+4=0$$ $$z=-\frac{4}{4y+7}$$
and setting $4y+7$ equal to $-4, -2, -1, 1, 2$ and $4$ gives the only solution $(2, -2, 4)$.
Toby Mak
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1It is quite hard to determine that $x=1$ and $x=2$ are the only possibilities for $x$. I'll now correct my comment: youcan show that if the expression lies between $1$ and $2$, then it must be equal to $\frac{3}{2}$, $\frac{4}{3}$, $\frac{5}{4}, \cdots, \frac{n+1}{n}$, and $\frac{10}{7}$ is not in this list. – Toby Mak Dec 05 '19 at 09:33