1. ## Prove~

how do we go about proving that all perfect squares have an odd number of divisors (other than 1 & themselves)?

how do we go about proving that all perfect squares have an odd number of divisors (other than 1 & themselves)?

Hint: if $n=p_1^{r_1} p_2^{r_2} \cdots p_k^{r_k}$ is the prime factorization of $n,$ then the number of all divisors of $n$ is $(r_1 + 1)(r_2 + 1) \cdots (r_k + 1).$

3. Originally Posted by NonCommAlg
Hint: if $n=p_1^{r_1} p_2^{r_2} \cdots p_k^{r_k}$ is the prime factorization of $n,$ then the number of all divisors of $n$ is $(r_1 + 1)(r_2 + 1) \cdots (r_k + 1).$
but how do you expect me to know this?

i have an exam on 10th of may. i have 2 papers. one with mcq's and one with proofs/ i can do all the mcqs but get stuck with proofs.

this came in the previous year (same exam) paper.

so how do we know these things?
(if you say practiceractice makes you all good at math & you can observe many such things with practice-but how would you know how to prove it?) and what if we dont? how are we supposed to do the problems then?

4. ## A visual "proof"

Adhyeta, here's a visual to show where it comes from:

Consider the divisors of $60=2^23^15^1$

$01=2^03^05^0$
$02=2^13^05^0$
$03=2^03^15^0$
$04=2^23^05^0$
$05=2^03^05^1$
$06=2^13^15^0$
$10=2^13^05^1$
$12=2^23^15^0$
$15=2^03^15^1$
$20=2^23^05^1$
$30=2^13^15^1$
$60=2^23^15^1$

Notice how every possible combination of exponents appears. For the exponent on 2, there are three choices: 0,1,2. For 3, there are two choices: 0,1. For 5, also two choices: 0,1. Going back to your basic probability, multiply these three together to get every possibility, hence $3*2*2=12$ divisors of 60.

This is not a formal proof, but it should show you that the equation NCA gave you did not come out of "thin air," but is rather the expression of a very simple observation that can be seen by factoring the divisors.

any divisor of $n$ has the same prime factors of $n.$ so it has to be in the form $d=p_1^{s_1} p_2^{s_2} \cdots p_k^{s_k},$ where $0 \leq s_j \leq r_j,$ for all $1 \leq j \leq k.$ so each $s_j$ has $r_j + 1$ possibilities. that's all!