Thread: Let n be an integer greater than one. Prove that n is a perfect square i.f.f. ...

1. Let n be an integer greater than one. Prove that n is a perfect square i.f.f. ...

Let n be an integer greater than one. Prove that n is a perfect square i.f.f. n's prime factorization, n=p_1*p_2*...*p_r, has only even exponents.

So I think I'm doing an analytical proof that establishes a biconditional with reference to the above statement.

By definition a perfect square is a number, n, such that n=m^2x=(m^x)^2 and after substituting,

I need to show that each exponent on a prime is even and I'm not entirely sure how to do this at the moment.

I don't think I can just go straight to

since I would need some kind of condition that requires each exponent to be an integer (maybe?) and that 2 divides any a with no remainder?

The proof of the other direction, from each prime factor having even exponents to n being a perfect square, is done already.

2. Re: Let n be an integer greater than one. Prove that n is a perfect square i.f.f. ...

Originally Posted by bkbowser
Let n be an integer greater than one. Prove that n is a perfect square i.f.f. n's prime factorization, n=p_1*p_2*...*p_r, has only even exponents.

So I think I'm doing an analytical proof that establishes a biconditional with reference to the above statement.

By definition a perfect square is a number, n, such that n=m^2x=(m^x)^2 and after substituting,

I need to show that each exponent on a prime is even and I'm not entirely sure how to do this at the moment.

I don't think I can just go straight to

since I would need some kind of condition that requires each exponent to be an integer (maybe?) and that 2 divides any a with no remainder?

The proof of the other direction, from each prime factor having even exponents to n being a perfect square, is done already.
In any prime factorization each exponent is an integer. If $\displaystyle p$ is prime then the $\displaystyle \sqrt p$ is irrational.

Does that not tell you that the exponents must be even?

3. Re: Let n be an integer greater than one. Prove that n is a perfect square i.f.f. ...

Originally Posted by Plato
In any prime factorization each exponent is an integer. If $\displaystyle p$ is prime then the $\displaystyle \sqrt p$ is irrational.

Does that not tell you that the exponents must be even?
I'm not sure I follow, or now that I think of it that this proof will work out, at least not the way I was thinking. I was hoping that the product of the of distinct primes and irrationals would be irrational but that's not true(?). Or at least way harder to show then I want to get into.

4. Re: Let n be an integer greater than one. Prove that n is a perfect square i.f.f. ...

OK I think I can work it out this way;

If n is a perfect square then it has the form of some integer, m, times itself.

By the Fundamental Theorem of Arithmetic n has a unique factorization into distinct primes up to order.

Since by definition a perfect square is some number times itself I have to be able to group the factors if n into two equal groups.

For any given a, if that a is even then it is perfectly divisible by two and so can be split into two equal groups.

For any given a, if that a is odd then it is not perfectly divisible into two groups. From which there are two cases;

Either you have integer exponents and unequal groupings, in which case A does not equal A and there is a contradiction with the definition of a perfect square.

Or, you have fractional exponents in equal groupings. In other words each grouping, A, contains at least one root and possibly more. Since each A is a grouping of distinct primes it is not that case that,

Where N is an integer.

Right here I want to say "As a perfect square is the square of an integer and as A can not be an integer, there is a contradiction if there is at least one odd exponentiation on a prime factor of n." I am however not entirely sure that the product of distinct primes and some variable number of roots of primes is not an integer.

Like the standard counter example to the claim that "The product of two irrational numbers is irrational" is that of,

which is of course not the case here since we're taking about the product of distinct primes with fractional exponents.

5. Re: Let n be an integer greater than one. Prove that n is a perfect square i.f.f. ...

If we have:

$\displaystyle n = m^2$

and:

$\displaystyle n = p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}$

(it is typical to insist that: $\displaystyle p_1 < p_2 < \cdots < p_k$ to avoid questions of "re-ordering" and also to leave out the special case n= 1),

then for each $\displaystyle p_j$ which is prime, we certainly have $\displaystyle p_j|n$ and then because $\displaystyle n = m\ast m$, and $\displaystyle p_j$ is PRIME, it divides one of m or m, that is to say, it divides m.

Since $\displaystyle p_j$ divides m, some highest positive power of it divides m. let's say this power is $\displaystyle (p_j)^t$. Thus:

$\displaystyle \frac{m}{(p_j)^t}$ is an integer, which $\displaystyle p_j$ does NOT divide.

Thus $\displaystyle \frac{n}{(p_j)^{2t}}$ is likewise an integer that $\displaystyle p_j$ does not divide.

By the uniqueness of the prime factorization for this integer, we have:

$\displaystyle \frac{n}{(p_j)^{2t}} = p_1^{r_1}\cdots (p_{j-1})^{r_{j-1}}(p_{j+1})^{r_{j+1}}\cdots p_k^{r_k}$

and cross-multiplication gives:

$\displaystyle n = p_1^{r_1}\cdots (p_{j-1})^{r_{j-1}}(p_j)^{2t}(p_{j+1})^{r_{j+1}}\cdots p_k^{r_k}$

so again by the uniqueness of prime factorization for n, we have:

$\displaystyle (p_j)^{2t} = (p_j)^{r_j}$

and uniqueness of factorization for THIS number forces:

$\displaystyle 2t = r_j$

that is, $\displaystyle r_j$ is even, for every j.