Exercise 2.1.42 (Same problem, $a,b,c$ in $\mathbb{Z}^*$)

Proof. Note that $a\equiv b(\mathit{mod}|c|)$ is equivalent to $a\equiv b(\mathit{mod}c)$. Now we search all solutions $(a,b,c)\in {(\mathbb{Z}\setminus \{0\})}^3$ of

Then $(\mathit{ka},\mathit{kb},\mathit{kc})$ is a solution for any integer $k$. Hence it suffices to determine all primitive triplets with the property $a\wedge b\wedge c=1$.

If $(a,b,c)$ is a solution, so is any permutation of $(a,b,c)$. So there is no loss in generality in assuming that $|a|\le |b|\le |c|$.

Note also that if $(a,b,c)$ is a solution, then $(-a,-b,-c)$ is also a solution, so we may suppose $c>0$.

Let $(a,b,c)$ be a solution of (1) such that $a\wedge b\wedge c=1$, $0<|a|\le |b|\le c$. We name such a solution a reduced solution.

Then (1) is equivalent to the existence of integers $k,l,m$ such that

Write $a={𝜀}_{a}A$, where ${𝜀}_a$ is the sign of $a$, and $A=|a|>0$. Similarly, $b={𝜀}_{b}B,B>0$, and $C=c>0$. Then $A\wedge B\wedge C=1$, $0<A\le B\le C$, and (1) is equivalent to

We study four cases, depending on the sign of $a$ and $b$.

• If $a>0,b>0$, then ${𝜀}_a=1,{𝜀}_b=1$. Then (3) is equivalent to

  $$\{\begin{array}{cc}\hfill B-A& =\mathit{kC},\hfill \\ \hfill C-B& =\mathit{lA},\hfill \\ \hfill C-A& =\mathit{mB}.\hfill \end{array}$$

  (4)

  Then Problem 41 shows that $(A,B,C)=(1,1,n)$, so

  $$(a,b,c)=(1,1,n),n>0.$$

• If $a<0,b<0$ then ${𝜀}_a=-1,{𝜀}_b=-1$. Then (3) is equivalent to

  $$\{\begin{array}{cc}\hfill B-A& =\mathit{KC},\hfill \\ \hfill C+B& =\mathit{LA},\hfill \\ \hfill C+A& =\mathit{MB},\hfill \end{array}$$

  (5)

  where $K=-k,L=-l,M=-m$ satisfy $K\ge 0,L>0,M>0$.

  From (5) we deduce $B=A+\mathit{KC}$, and $C+A=M(A+\mathit{KC})$, thus

  $$\begin{array}{lll}\hfill 0=(M-1)A+(\mathit{MK}-1)C.& & \hfill \text{(6)}\end{array}$$

  • If $K=0$, then

    $$\begin{array}{llll}\hfill B& =A,& \hfill & \\ \hfill C& =(L-1)A.& \hfill & \end{array}$$

    Since $A\wedge B\wedge C=1$, $A=1$, thus $(A,B,C)=(1,1,n),n>0$, so

    $$(a,b,c)=(-1,-1,n),n>0.$$

  • If $K\ge 1$, then $M-1\ge 0$ and $\mathit{MK}-1\ge 0$, so (6) shows that $M=1,\mathit{KM}=1$ (otherwise $0=(M-1)A+(\mathit{MK}-1)C>0$), thus $M=K=1$. In this case, (5) is equivalent to

    $$\{\begin{array}{cc}\hfill B+C& =\mathit{LA},\hfill \\ \hfill B-C& =A,\hfill \end{array}$$

    (7)

    but this is impossible, because $B-C=A$ implies $B>C$, but $B\le C$ for a reduced solution.

• If $a<0,b>0$, then ${𝜀}_a=-1,{𝜀}_b=1$, so (3) gives

  $$\{\begin{array}{cc}\hfill B+A& =\mathit{KC},\hfill \\ \hfill C-B& =\mathit{LA},\hfill \\ \hfill C+A& =\mathit{MB},\hfill \end{array}$$

  (8)

  where $K=k,L=-l,M=m$, and $K>0,L\ge 0,M>0$.

  From (8) we deduce $C=B+\mathit{LA}$, and $B+A=K(B+\mathit{LA})$, thus

  $$\begin{array}{lll}\hfill 0=(K-1)B+(\mathit{KL}-1)A.& & \hfill \text{(9)}\end{array}$$

  • If $L=0$, then $B=C$, and (8) shows that $A=(K-1)C\ge C$, thus $A=C$, so that $A=B=C$. Since $A\wedge B\wedge C=1$, $A=B=C=1$, so

    $$(a,b,c)=(-1,1,1).$$

  • If $L\ge 1$, then $K-1\ge 0,\mathit{KL}-1\ge 0$. The equation (9) shows that $K=L=1$. Thus (8) is equivalent to

    $$\{\begin{array}{cc}\hfill C-A& =B,\hfill \\ \hfill C+A& =\mathit{MB},\hfill \end{array}$$

    (10)

    which gives

    $$\{\begin{array}{cc}\hfill 2A& =(M-1)B,\hfill \\ \hfill 2C& =(M+1)B,\hfill \end{array}$$

    (11)

    • If $M$ is odd, then (11) gives $A=(n-1)B,C=\mathit{nB}$, where $n=(M+1)∕2$ is a positive integer. Since $A\wedge B\wedge C=1$, we obtain $(A,B,C)=(n-1,1,n)$. But the condition $0<A\le B\le C$ shows that $n=2$, thus $(A,B,C)=(1,1,2)$, and

      $$(a,b,c)=(-1,1,2).$$

    • If $M$ is even, then $2\mid B$, and

      $$\{\begin{array}{cc}\hfill A& =(M-1)(B∕2),\hfill \\ \hfill C& =(M+1)(B∕2),\hfill \end{array}$$

      (12)

      Since $A\wedge B\wedge C=1$, $B∕2=1$, and $(A,B,C)=(M-1,2,M+1),M>0$. The condition $0<A\le B\le C$ gives $M=2$ or $M=3$, but $M$ is even, thus $(A,B,C)=(1,2,3)$.

      $$(a,b,c)=(-1,2,3).$$

• If $a>0$ and $b<0$, then ${𝜀}_a=1,{𝜀}_b=-1$, so (3) gives

  $$\{\begin{array}{cc}\hfill B+A& =\mathit{KC},\hfill \\ \hfill C+B& =\mathit{LA},\hfill \\ \hfill C-A& =\mathit{MB},\hfill \end{array}$$

  (13)

  where $K=-k,L=l,M=-m$, and $K>0,L>0,M\ge 0$. From (13) we deduce $C=A+\mathit{MB}$, and $B+A=K(A+\mathit{MB})$, thus

  $$0=(K-1)A+(\mathit{KM}-1)B.$$

  (14)

  • If $M=0$, then $A=C$, and (12) gives $B=(K-1)C$. Since $A\wedge B\wedge C=1$, we obtain $A=1,B=n,C=1$, where $n=K-1>0$. The condition $0<A\le B\le C$ gives $n=1$, so

    $$(a,b,c)=(1,-1,1).$$

  • If $M\ge 1$, then $K-1\ge 0,\mathit{KM}-1\ge 0$. The equation (14) shows that $K=M=1$, so (12) is equivalent to

    $$\{\begin{array}{cc}\hfill C-B& =A,\hfill \\ \hfill C+B& =\mathit{LA},\hfill \end{array}$$

    (15)

    which gives

    $$\{\begin{array}{cc}\hfill 2B& =(L-1)A,\hfill \\ \hfill 2C& =(L+1)A,\hfill \end{array}$$

    (16)

    • If $L$ is odd, then $B=(n-1)A,C=\mathit{nA}$, where $n=(L+1)∕2>0$. Since $A\wedge B\wedge C=1$, $A=1$, and $(A,B,C)=(1,n-1,n)$, where $1\le n-1$, therefore

      $$(a,b,c)=(1,1-n,n),n\ge 2.$$

    • If $L$ is even, $L=2n$ for some $n>0$. Then $A$ is even, and

      $$\{\begin{array}{cc}\hfill B& =(L-1)(A∕2),\hfill \\ \hfill C& =(L+1)(A∕2).\hfill \end{array}$$

      (17)

      Since $A\wedge B\wedge C=1$, $A∕2=1$, and $(A,B,C)=(2,L-1,L+1)=(2,2n-1,2n+1)$, where $2\le 2n-1$, thus

      $$(a,b,c)=(2,-2n+1,2n+1),n\ge 2.$$

To conclude, the reduced solutions, i.e. the solutions $(a,b,c)$ such that $0<|a|\le |b|<c$, are

(Same results as in “Answers”, p. 513; $(1,-1,2)$ is given by $(1,1-n,n)$ for $n=2$.)

We obtain all solutions by taking all permutations of these solutions, and multiplying them by some non zero constant. □