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So this is a homework question, and I'm stuck. I'm also, to be perfectly honest, not quite sure where to start.

What I tried to do is take the square root of both $40$ and $a^2$ and I got $2\sqrt{10}\text{ } | \text{ }a$ and from here I could see that if I factored 40 before I took the square root and forced the result, then I could get $(\sqrt{4} * 10) | \sqrt{a^2}$ which would give me $20|a$, but that seems, to me, to be mathematically incorrect, as I would assume that I have to take the square root of everything...I also know that I can't just assume that $a^2$ would be divisible by $2$ as that was my next thought, but $9^2 = 81$ and so dividing by 2 just because seemed like an even worse idea...

I know I haven't really tried (with any success) anything to solve this, but any help would be greatly appreciated!

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    $\begingroup$ Notice that $40 = 2^3 \cdot 5$, so can you show that we must have $2^2 | a$ and $5 | a$? $\endgroup$
    – Krijn
    Commented Oct 4, 2015 at 22:12
  • $\begingroup$ $40|a^2$ means that $ k=a^2/40 \in Z$ .But $k$ need not be the square of an integer. So $\sqrt 40 | a $, which means that $\sqrt k=a/\sqrt 40 \in Z$ , may be false. For example $a=100, k=250, a/\sqrt 40=5\sqrt 10\not \in Z$. $\endgroup$ Commented Oct 4, 2015 at 23:30

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Hint: If $40\mid a^2$, then $10\mid\left(\frac a2\right)^2$

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Hint: can you prove that $a$ is even? Divisible by $4$? Divisible by $5$? Do it step by step and put the pieces together.

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    $\begingroup$ Thanks for the hints, I've solved it (I think). I said that $(2^3 \times 5) | a^2$ and since $a^2 = a \times a$, each prime factor of $a$ must be included in $a^2$ an even number of times. Thus we know that $a^2$ has at least $ 2^2 \times 2^2 \times 5^2$ as a factor. If I take the square root of both sides I'm left with $ (2 \times 2 \times 5) | a$ and so $20 |a$ Does that make sense? Or did I make a weird leap somewhere? $\endgroup$
    – Rebecca
    Commented Oct 4, 2015 at 22:26
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    $\begingroup$ @Rebecca, that is correct. $\endgroup$
    – Krijn
    Commented Oct 4, 2015 at 22:34
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I'll try a proof by contradiction.

Suppose $20 \not\mid a$. Then either $4\not\mid a$ or $5\not\mid a$.

If $4 \not\mid a$, then $a =4k+j $, where $1 \le j \le 3$. Then $a^2 =16k^2+8kj+j^2 =8(2k^2+kj)+j^2 $. But, $8 \not\mid j^2$ so $8 \not\mid a^2$, so $40 \not\mid a^2$.

Similarly, if $5 \not\mid a$, then $a = 5k+j$ where $1 \le j \le 4$. Then $a^2 =25k^2+10kj+j^2 =5(5k^2+2kj)+j^2 $. But $5 \not\mid j^2$, so $5 \not\mid a^2$, so $40 \not\mid a^2$.

Therefore, if $20 \not\mid a$, then $40 \not\mid a^2$, and we are done.

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i think i have a solution.

$40=2^3\times 5|a^2=a\times a$.$\space$ therefore $2|a^2$ so $2|a$.$\space$ also, $5|a$.$\space$ so $a=(5b)(2c);\space b,c\in Z^+$.$\space$ making $a^2=5^22^2b^2c^2,$ where $2|b^2c^2$ because $2^3|a^2.$$\space$ therefore $2|b^2$ or $2|c^2$ so $2|b$ or $2|c$.$\space$ so $b^2c^2 =(2^2{p_2}^2{p_3}^2...{p_n}^2)(b^2$ or $c^2)$;$\space p_2,p_3,...,p_n$ other prime factors,$\space n\in Z^+$.

so $a^2=5^22^2(2^2{p_2}^2{p_3}^2...{p_n}^2)(b^2$ or $c^2)$; and $a=5^12^2\sqrt{{p_2}^2{p_3}^2...{p_n}^2(b^2\space or\space c^2)}=20(\sqrt{{p_2}^2{p_3}^2...{p_n}^2(b^2\space or\space c^2)})$. $\space \therefore 20\mid a$

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  • $\begingroup$ What does it mean for $4\sqrt5\mid a$ when $4\sqrt5\not\in\mathbb{Z}$? $\endgroup$
    – robjohn
    Commented Oct 4, 2015 at 23:19
  • $\begingroup$ ah i see . my mistake $\endgroup$
    – miniparser
    Commented Oct 4, 2015 at 23:23
  • $\begingroup$ it has been edited. $\endgroup$
    – miniparser
    Commented Oct 5, 2015 at 1:01

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