If you multiply a square number with a non square number, the result is never a square number.
Here, a square number is product of an integer with itself.
Do you agree with this statement? If so, explain why. If you disagree with this statement, please give me an counterexample where the answer is a square number.
On
It follows easily be the Fundamental Theorem of Arithmetic, but it can be done with weaker tools. Namely if for integers $\, a^2 = b^2 c \,$ for a nonsquare $\,c\,$ then $\,x^2 - c\,$ has a rational root $\,x = a/b\,$ so, by the Rational Root Test, $\,a/b = d\,$ is an integer, so $\, c = (a/b)^2 = d^2,\,$ contra hypothesis.
Remark $\ $ As above, the problem is equivalent to the standard irrationality proofs of square-roots, about which there are many prior threads here, e.g. this one.
Note that this property may fail for other numbers, e.g. in $\,\Bbb Z[\sqrt{12}]\,$ we have $\, 3\cdot 2^2 = \sqrt{12}^2$ therefore the product of a nonsquare and a square is a square for some quadratic numbers. Thus any proof must necessarily employ some property special to the ring of integers, in particular, some property not shared by all quadratic number rings, e.g. unique factorization; Euclid's Lemma.
Using the Fundamental Theorem of Arithmetic we can see if $a^2=b$ and the prime factorization of $a$ is $2^{a_2}\cdot3^{a_3}\cdot5^{a_3} \dots$ then the prime factorization of $a^2$ is $2^{2a_2}\cdot3^{2a_3}\cdot5^{2a_3} \dots$ So this tells us the prime factorization of squares has even exponents. Let k be a square and m a non-square. Then $mk$ has some odd exponents and therefore is not a square.