Exercise 1.3.22 (Q.C.M.)

Answers:

(1)

“If $a\wedge b=a\wedge c$ then $a\vee b=a\vee c$.”

FALSE.
Counterexample: $a=2,b=8,c=4$. Then $a\wedge b=a\wedge c=2$, but $a\vee b=4\ne a\vee c=8$.

(2)

“If $a\wedge b=a\wedge c$ then $a^2\wedge b^2=a^2\wedge c^2$.”

TRUE.
Proof: If $d=a\wedge b=a\wedge c$, by Problem 18, $d^2=a^2\wedge b^2=a^2\wedge c^2$.

(3)

“If $a\wedge b=a\wedge c$ then $a\wedge b=a\wedge b\wedge c$.”

TRUE.
Proof: Put $d=a\wedge b\wedge c$. Then

• $d\ge 0;$
• $d\mid a$ and $d\mid b$;
• For any integer $\delta$, if $\delta \mid a$ and $\delta \mid b$, then $\delta \mid a\wedge b$. Since $a\wedge b=a\wedge c$, $\delta \mid a\wedge c$. Therefore $\delta \mid a,\delta \mid b$ and $\delta \mid c$. Hence $\delta \mid a\wedge b\wedge c=d$.

By the characterization of the gcd (Theorem 1.4), this shows that $a\wedge b=d=a\wedge b\wedge c$.

(4)

“If $p$ is a prime and $p\mid a$ and $p\mid a^2+b^2$ then $p\mid b$.”

TRUE.
Proof: From $p\mid a$, we deduce $p\mid a^2$. Then $p\mid (a^2+b^2)-a^2$, thus $p\mid b^2$. Since $p$ is prime, $p\mid b$.

(5)

“If $p$ is a prime and $p\mid a^7$ then $p\mid a$.”

TRUE.
Proof: By Theorem 1.15, where $a_1=a_2=\cdots =a_7=a$, if $p\mid a^7$, then $p\mid a$.

(6)

“If $a^3\mid c^3$, then $a\mid c$.”

TRUE.
Let $d=a\wedge c$. By (generalized) Problem 18, $d^3=a^3\wedge c^3$. If $a^3\mid c^3$, then $d^3=a^3\wedge c^3=a^3$, thus $a=d=a\wedge c$, so $a\mid c$.

(7)

“If $a^3\mid c^2$, then $a\mid c$.”

TRUE.
Proof: If $a^3\mid c^2$, a fortiori $a^2\mid c^2$. As in part (6), $a\mid b$

(8)

“If $a^2\mid c^3$, then $a\mid c$.”

FALSE.
Counterexample: $a=8,c=4$: $8^2\mid 4^3$ but $8\nmid 4$.

(9)

“If $p$ is a prime and $p\mid (a^2+b^2)$ and $p\mid (b^2+c^2)$ then $p\mid (a^2-c^2)$.”

TRUE.
Proof: $p\mid (a^2+b^2)-(b^2+c^2)=a^2-c^2$.

(10)

“If $p$ is a prime and $p\mid (a^2+b^2)$ and $p\mid (b^2+c^2)$ then $p\mid (a^2+c^2)$.”

FALSE.
Counterexample: $5\mid 1^2+2^2$ and $5\mid 2^2+4^2$, but $5\nmid 1^2+4^2$.

(11)

“If $(a,b)=1$ then $(a^2,\mathit{ab},b^2)=1$.”

TRUE.
Proof: If $a\wedge b=1$, then $a^2\wedge b^2=1$. A fortiori $a^2\wedge \mathit{ab}\wedge b^2=1$.

(12)

“ $a^2\vee \mathit{ab}\vee b^2=a^2\vee b^2$.”

TRUE.
Proof: For all integers $r,s$, $\max <!--\; FUNCTION\; APPLICATION\; -->(2r,r+s,2s)=\max <!--\; FUNCTION\; APPLICATION\; -->(2r,2s)$. Indeed, if $r\le s$, then $2r\le r+s\le 2s$, and $\max <!--\; FUNCTION\; APPLICATION\; -->(2r,r+s,2s)=2s=\max <!--\; FUNCTION\; APPLICATION\; -->(2r,2s)$. By symmetry, we have the same result if $r\ge s$. Therefore, if $a={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{p\in A}{p}^{\alpha (p)},b={\prod }_{p\in A}{p}^{\beta (p)},$ then

$$\begin{array}{llll}\hfill a^2\vee \mathit{ab}\vee b^2& =\underset{p\in A}{\prod }{p}^{\max <!--\; FUNCTION\; APPLICATION\; -->(2\alpha (p),\alpha (p)∕\beta (p),2\beta (p))}& \hfill & \\ \hfill & =\underset{p\in A}{\prod }{p}^{\max <!--\; FUNCTION\; APPLICATION\; -->(2\alpha (p),2\beta (p))}& \hfill & \\ \hfill & =a^2\vee b^2.& \hfill & \end{array}$$

(13)

“If $b\mid (a^2+1)$ then $b\mid (a^4+1)$.”

FALSE.
Counterexample: $a=2,b=5$, since $5\mid 2^2+1$ but $5\nmid 2^4+1$.

(14)

“If $b\mid (a^2+1)$ then $b\mid (a^4-1)$.”

TRUE.
Proof: If $b\mid (a^2+1)$ then $b\mid (a^2-1)(a^2+1)=a^4-1$.

(15)

“ $a\wedge b\wedge c=(a\wedge b)\wedge (a\wedge c)$.”

TRUE.
Proof: $(a\wedge b)\wedge (a\wedge c)=a\wedge b\wedge a\wedge c=(a\wedge a)\wedge b\wedge c=a\wedge b\wedge c$.