This problem is taken from An Introduction to The Theory Of Numbers, by Ivan Niven.
Find all such pairs $a,b,c$ such that $a\equiv b \pmod c$, $b\equiv c \pmod a$,
$c\equiv a \pmod b$. I wrote $a=b+ck_1,\ b=c+ak_2, c=a+bk_3$ and equated the determinant to $0$, but didn't get anything much useful out of it.
Find all such pairs a,b,c such that $a\equiv b \pmod c$, $b\equiv c \pmod a$,$c\equiv a \pmod b$
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elementary-number-theory
congruences
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0I have edited your equations $a=b+ck_1,\ldots$. I hope now that is what you meant. – 2017-01-26
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0If $a,b,c$ are positive then it can be shown that $(a,b,c)=(tb,b,b)$ I believe. ($t$ is an integer). – 2017-01-26
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0@EclipseSun Yes, Thank You – 2017-01-26
1 Answers
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In order that the homogeneous system
$$\begin{cases}\ \ \ \ a&-&b&-&k_1c&=&0\\-k_2a&+&b&-&c&=&0\\ \ -a&-&k_3b&+&c&=&0\end{cases}$$
has a solution different from the trivial solution $(0,0,0)$, we must have:
$$\tag{1}\begin{vmatrix}1&-1&-k_1\\-k_2&1&-1\\-1&-k_3&1\end{vmatrix}=0 \ \ \iff \ \ k_1k_2k_3+k_1+k_2+k_3=0$$
Either one of the $k_i=0$ (giving for example when $k_1=0$, $k_2$ arbitrary and $k_3=-k_2$), or we can write $(1)$ under the form:
$$\tag{2}\iff \dfrac{1}{k_2k_3}+\dfrac{1}{k_3k_1}+\dfrac{1}{k_1k_2}=-1$$
and there are very few integers triples $(k_1,k_2,k_3)$ that verify $(2)$.
Up to you...
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1What if $k_1=0$? And your equation is wrong. It should be $\frac{1}{k_{1}k_{2}}$. – 2017-01-26
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0I already did this. What do I do after this? – 2017-01-26
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0@S.C.B Right. I correct it. – 2017-01-26
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0@bat_of_doom You already had obtained the last equality ? Very well. Then the observation is that there are not many integer triples that very this identity... – 2017-01-26
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0@JeanMarie Why did you re-arrange it in the last form? Is there a shorter way out? I equated $k_3=-\frac{k_1+k_2}{1+k_1k_2} \implies k_2\equiv 1 \pmod {k_1}, k_1 \equiv 1 \pmod {k_2}.$ – 2017-01-26
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0The implication you give is not evident. With the last form I give, the discussion is about the fact that, once you have eliminated the 3 cases where one of the $k_i=1$ and the two others are equal to $-1$, it remains very few cases... – 2017-01-26