Two integers and thus their squares have no common factors

[Happiness]

But of course, I have not seen your full proof, which might clear this issue up.

Let $m=p_1^{k_1}p_2^{k_2}...p_n^{k_n}$. Given that $m$ is not a square number, there exist at least one $k_n$ that is odd. Suppose we call them $k_j$'s and denote those even $k_n$'s as $k_i$'s. Then for $\sqrt{m}$ to be rational, $\sqrt{p_1^{k_1}p_2^{k_2}...p_n^{k_n}}$ has to be rational. $\sqrt{p_i^{k_i}}$ are integers and hence rational. So $\sqrt{\Pi_jp_j^{k_j}}$ and hence (we are using $k_j=1$ mod 2) $\sqrt{\Pi_jp_j}$ has to be rational. (But it is not rational, so we [will] have a contradiction.)

In a few more steps, we can show that this leads to a contradiction.

The $p_n$'s are prime numbers while the $k_n$'s are integers, of course.

 

[PeroK]

Let $m=p_1^{k_1}p_2^{k_2}...p_n^{k_n}$. Given that $m$ is not a square number, there exist at least one $k_n$ that is odd. Suppose we call them $k_j$'s and denote those even $k_n$'s as $k_i$'s. Then for $\sqrt{m}$ to be rational, $\sqrt{p_1^{k_1}p_2^{k_2}...p_n^{k_n}}$ has to be rational. $\sqrt{p_i^{k_i}}$ are integers and hence rational. So $\sqrt{\Pi_jp_j^{k_j}}$ and hence $\sqrt{\Pi_jp_j}$ has to be rational. (But it is not rational, so we [will] have a contradiction.)

In a few more steps, we can show that this leads to a contradiction.

My proof would be:

Let $q = \frac{a}{b}$ be a proper rational (where $a,b$ have no common factors and $b \ne 1$). Now $a^2$ and $b^2$ have the same prime factors as $a$ and $b$ respectively, hence have no common factors. And, as $b^2 \ne 1$, $q^2 = \frac{a^2}{b^2}$ is a proper rational.

Hence, a proper rational squares to a proper rational, never to a whole number.

Therefore, the square root of a whole number is either a whole number or irrational (never a rational).

 

[Happiness]

Therefore, the square of a whole number is either a whole number or irrational (never a rational).

I suppose you mean "Therefore, the square root of a whole number is either a whole number or irrational (never a rational)."

 

[PeroK]

I suppose you mean "Therefore, the square root of a whole number is either a whole number or irrational (never a rational)."

Yes, fixed now!

 

[Erland]

PeroK's proof is the best one, in my opinion.

 

[bahamagreen]

not a mathematician (like you couldn't tell...) but here is my attempt...

let p<q, no common factor
if p is even (two is a factor) then q cannot be even
if p is threeven (three is a factor) then q cannot be threeven
if p is fourven then p is also even (an integer sub-multiple of four), so q cannot be either

so generally if p is nven (n is a factor) then q cannot be nven nor any integer sub-multiples of n of nven
q cannot be xp nor p^x where x = int > 1; where x=1 then p is pven (p is a factor) so q cannot be pven

for p and q to share a common factor n entails that n<p
but all nven of p for n on q up to p yield no common factor

so for p^2 and q^2 to be nven (share a common factor n),
then p>n<p^2 is the only range to look for n since all nven of p on q where n</=p are eliminated
but all n>p where n=xp<p^2 are already eliminated as well, including all x from 1 to p (where np=pp=p^2)
therefore p^2 and q^2 have no common factor

 

[Happiness]

not a mathematician (like you couldn't tell...) but here is my attempt...

let p<q, no common factor
if p is even (two is a factor) then q cannot be even
if p is threeven (three is a factor) then q cannot be threeven
if p is fourven then p is also even (an integer sub-multiple of four), so q cannot be either

so generally if p is nven (n is a factor) then q cannot be nven nor any integer sub-multiples of n of nven
q cannot be xp nor p^x where x = int > 1; where x=1 then p is pven (p is a factor) so q cannot be pven

for p and q to share a common factor n entails that n<p
but all nven of p for n on q up to p yield no common factor

so for p^2 and q^2 to be nven (share a common factor n),
then p>n<p^2 is the only range to look for n since all nven of p on q where n</=p are eliminated
but all n>p where n=xp<p^2 are already eliminated as well, including all x from 1 to p (where np=pp=p^2)
therefore p^2 and q^2 have no common factor

What do you mean by "nven of p for n on q"? Could you give an example?

 

[bahamagreen]

"but all nven of p for n on q up to p yield no common factor"

nven is divisibility by n; for p where nven is true, it cannot be true for q

"nven of p" is the case of p divisible by n being true e.g., p is "even" (AKA "twoven") and divisible by 2
"for n" is the factor number for nven e.g., "even" means 2 is a factor
"on q" means q is the object for which this factor cannot apply e.g., therefore q is not even
"up to p" means the n of nven may take integer values from 2 up to but not including p e.g., even (twoven), threeven, fourven... up to but not including pven (because p is a factor)

sorry for the terrible wording and made up terms...

edit - My first intuition about this problem was to assign p and q as y and x intercepts on the axes and wonder if there was anything special that distinguished the lines through p and q (and subsequently the lines through p^2 and q^2) when p and q had vs did not have common factors... but I think that may be similar to what is going on when the various approaches are using p/q?

 

[Happiness]

so for p^2 and q^2 to be nven (share a common factor n),
then p>n<p^2 is the only range to look for n since all nven of p on q where n</=p are eliminated
but all n>p where n=xp<p^2 are already eliminated as well, including all x from 1 to p (where np=pp=p^2)
therefore p^2 and q^2 have no common factor

The third line is not explained clearly. You have only established that $q$ is not divisible by $p$ (since they don't have a common factor). But it has not been established that $q^2$ is not divisible by $p$, which is your deduction in the third line.

For example, $q=30$ is not divisible by $p=12$, but $q^2=900$ is. So it's not clear how you come to the deduction that $q^2$ is not divisible by $p$.

Also, in the second line where you deduce that $q^2$ and $p^2$ have no common factor $n$ for $n\leq p$ because $q$ and $p$ have no common factor $n$ for $n\leq p$, it looks like an inverse error is committed.

It's better to denote the common factor of $p^2$ and $q^2$ as $m$, to distinguish it from $n$, the common factor of $p$ and $q$.