0

Need prove that : $$d\mid a \iff d^2\mid a^2$$

There are two parts: if, and only if. It also means that from both directions the other side must be provable.

My approach is to take only two categories of numbers as even or odd. Now, the even is denoted by $2n$, for $n$ being a natural (generalizing to integers can be easily done).

Now, $d^2$ must be also even, and if $d\mid a$ then a must also be even.

Similarly, for odd numbers.

drhab
  • 153,781
jiten
  • 4,960
  • For proving the inverse you could use prime-factorization $d=p_1^{r_1}\cdots p_k^{r_k}$ where the $p_i$ are distinct primes and the $r_i$ are positive integers. – drhab Nov 10 '17 at 08:16
  • @drhab Proving the inverse using the prime-factorization of the divisor, and then taking the square of that is confusing me. I hope you meant to find prime-factorization of $p^2$ instead. Now, taking the square root means simply halving the powers of all individual primes. How it will it help in proving the inverse is confusing still, as may be need have prime-factorization of $a^2$ too; and halving the powers of its constituent primes in the corresponding prime-factorization. But, then what will it lead to is still unclear to me. Please elaborate more. – jiten Nov 10 '17 at 11:33

2 Answers2

1

Let it be that $d=p_1^{r_1}\times\cdots\times p_k^{r_k}$ where the $p_i$ are distinct primes and the $r_i$ are positive integers.

Now suppose that $d$ does not divide $a$.

Then for at least one $i\in\{1,\dots,k\}$ we will have $\neg (p_i^{r_i}\mid a)$.

Apparantly - if $p_i^{s_i}\mid a$ with maximal $s_i$ - we have $s_i<r_i$.

A direct consequence of this is that $\neg (p^{2r_i}\mid a^2)$, since we also have $2s_i<2r_i$.

Then $\neg (d^2=p_1^{2r_1}\times\cdots\times p_k^{2r_k}\mid a^2)$

Proved is now that: $$\neg(d\mid a)\implies\neg(d^2\mid a^2)$$ or equivalently:$$d^2\mid a^2\implies d\mid a$$

drhab
  • 153,781
  • My elaboration of your answer is: For the case when $d \nmid a$, there are 2 disjoint sub-cases - (i) some prime factor (of $d$) does not exist in prime-factorization of $a$, (ii) if there are all common prime factors, then the power of that (some, $\ge 1$) prime factor(s) is lesser in $a$, as compared to $d$. I badly need your vetting, as feel I have made at least some error. – jiten Nov 10 '17 at 12:58
  • That is a correct interpretation. Not being present of the prime $p_i$ in $a$ is the same as $s_i=0 <r_i$ – drhab Nov 10 '17 at 13:08
0

To use cases in odd and even is often a good start in excercises in number theory like this one. However in this case it is not necessary. Just use the definition for divisibility.

$d | a$ iff there is a number $k$ such that $dk = a$.

Using this we can conduct the following calculations $$d|a \Leftrightarrow dk=a \Leftrightarrow dk*dk=a*a \Leftrightarrow d^2*k^2=a^2 $$ Thus if we let $k_0 =k^2$ we see that $d^2*k_0=a^2$ which, again by the definition of dividing implies that $d^2 | a^2$.

Regarding the consistency of your proof, It kinda proves (or at least stipulate) that $d^2$ is even and that $a$ is even... but why would this imply that $d^2 | a^2?$ it actually only prove that $2|a^2$.

Edit: Indeed, this does only prove one of the directions. I'm sorry for that, I'll come back, however the second direction is much more elaborate, and I'm not sure how to do it without using a lot more theory...

Ove Ahlman
  • 4,339
  • Thanks, but the reverse or the only-if condition is also requested. I can only pursue so far as: if $d^2 | a^2$; then taking the positive square root of both will lead to d|a; and assuming that the if part (as shown by you) is already proved. – jiten Nov 10 '17 at 08:05
  • This doesn't actually do the reverse direction. You've shown that if $a^2=d^2k^2$ for some integer $k$ then $d\mid a$. There's a missing step: you need to show that if $a^2=d^2j$ for some integer $j$ then $j=k^2$ for some integer $k$. – Especially Lime Nov 10 '17 at 08:16
  • @EspeciallyLime If I have to show that $j=k^2$ for some integer k; then it only proves that $k^2$ has a natural as square root. How to show it is another question though, and am not clear about it. – jiten Nov 10 '17 at 11:39