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To prove the statement above, I made this attempt:

Let a, b, c ∈ ℤ, and a | b and a | c. Then ∃ j, k ∈ ℤ such that aj = b and ak = c. Then

$b^2$ + $c^2$ = $(aj)^2$ + $(ak)^2$ = $a^2$($j^2$ + $k^2$).

The problem is I need to show a(integer), not $a^2$. If I remove the power from a, I get a $\sqrt{(j^2 + k^2)}$, which is not necessarily an integer,

Any idea what I'm missing?

Thank you!

Val
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  • if $a|b$ and $a|c,$ then $a|b\times b $ and $a|c\times c,$ so $a|b^2+c^2$ – J. W. Tanner Dec 16 '19 at 03:06
  • I have to start with $^2$ + $^2$ and arrive at a(integer). – Val Dec 16 '19 at 03:08
  • The multiples of $a$ are closed under addition and integer scalings so closed under integral linear combinations, so if $b,c$ are multiples so too is $,nb+mc,$ for all $,n,m\in \Bbb Z.,$ OP is special case $n=b,m=c.\ $ See here for more. – Bill Dubuque Dec 16 '19 at 07:27

1 Answers1

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$$ a^2(j^2 + k^2) = a \left( a(j^2 + k^2) \right) \text{.} $$

( I probably would have gone with \begin{align*} \frac{b^2 + c^2}{a} &= \frac{b^2}{a} + \frac{c^2}{a} \\ &= \left( \frac{b}{a} \right)b + \left( \frac{c}{a} \right) c \\ & \in \mathbb{Z} \text{.)} \end{align*}

Eric Towers
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  • That makes sense. But If I were to stick to (($^2$ + $^2$)), would that be correct? Prof. noted that I have to arrive to a(integer). – Val Dec 16 '19 at 03:18
  • @Val : You have stated $j,k \in \Bbb{Z}$. The integers are closed under multiplication, so $j \cdot j \in \Bbb{Z}$ and $k \cdot k \in \Bbb{Z}$. The integers are closed under addition, so $j^2 + k^2 \in \Bbb{Z}$ and then, since you have stated $a \in \Bbb{Z}$, multiplication by $a$ gives $a(j^2 + k^2) \in \Bbb{Z}$, which is what you already have... – Eric Towers Dec 16 '19 at 03:31
  • I see now, I need a kick in the head. Thank you! – Val Dec 16 '19 at 03:40