Product of 2 squarefree integers

Question

Prove that the product of any 2 distinct squarefree integers results in an integer that is not a square. Source: My math teacher

I know this is true because when you prime factorize 2 arbitrary squarefrees and then multiply, there will always be an extra prime in the product. I just don't know how to write this formally.

Answer 1

A formal proof would proceed as follows:

Let $a = a_1 a_2 \cdots a_n$ and $b = b_1 b_2 \cdots b_m$ be the two square-free integers. As they are not equal, there is at least one factor that does not appear in both numbers. Without loss of generality, you can assume this is $a_1$. Now if you look at $ab = \prod_{i = 1}^n a_j \prod_{j=1}^m b_j$, then you see that the exponent of $a_1$ is at most $1$, hence it can't be a perfect square, as it would have to have even exponent for each prime factor that appears in it.

Answer 2

Your idea is fine indeed for a squarefree integers the factorization is in the form

$$q=\prod_i p_i$$

and since the factorization is unique, two distinct squarefree integers must present at least two distinct primes in their factorizations and from there it is easy to conclude.

Notably we have

$$q_1=p_{1,0}\prod_i p_{1,i}\quad q_2=p_{2,0}\prod_i p_{2,i} \implies q_1q_2=p_{1,0}p_{2,0}\prod_i p_{3,i}$$

which is not a square.

Answer 3

For a more unconventional proof, use nimbers.

Each squarefree positive integer may be associated with a number according to the mapping:

$2^a×3^b×5^c×...\to (a×2^0)+(b×2^1)+(c×2^2)+...$

where by definition each exponent $a,b,c,...$ is $0$ or $2$.

Then we find the square-free core of the product by applying nimber or exclusive-or addition to the nimbers. For instance, $10=2×5$ gives a nimber of $1+0+4=5$ and $15=3×5$ gives $0+2+4=6$, the nimber sum is $1+2+0=3$ and this corresponds to the squarefree core of $10×15$ equalling $2×3=6$. We get a square iff the squarefree core is $1$, whose corresponding nimber is $0$, but for two nimbers a zero sum is obtained only if the input nimbers are identical and therefore only if the square-free integers with which we started are identical.

The nimber concept may be generalized to products of more integers. In the above scheme the nimbers corresponding to $6,10,15$ are respectively $3,5,6$ and $3\oplus5\oplus6=0$. So $6×10×15$ will be a square ($30^2$).