Square numbers multiplied by non-square numbers

Question

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.

@fkraiem What are you thinking here? I can't think of any reasonable definition of "numbers" where the answer would make a difference. — MJD, Mar 12 '14 at 20:40
@user135034, have you looked at the link I provided? It's the exact same question! — recursive recursion, Mar 12 '14 at 20:51
@MJD One should think a little bit harder, 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. — Bill Dubuque, Mar 12 '14 at 21:37
@MJD Thanks. TIP: one can often find counterexamples like these in simple quadratic number rings that lack unique factorization. — Bill Dubuque, Mar 12 '14 at 21:51
How come literally no one even thought about, or at least mentioned, the edge case ($0$ times any nonsquare equals $0 = 0^2$ is a square)? — Najib Idrissi, Jan 12 '15 at 08:46

Asinomás · Answer 1 · 2014-03-12T22:05:25.960

3

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.

edited Mar 12 '14 at 22:05

answered Mar 12 '14 at 20:52

Asinomás

105,651

This answer has the right idea, however the prime number theorem is not the theorem you are thinking of. – Ben Millwood Mar 12 '14 at 21:19
(You're looking for this ostentatiously-named theorem of arithmetic) – Ben Millwood Mar 12 '14 at 21:29
Oh, of course, thanks for pointing it out. ;) – Asinomás Mar 12 '14 at 22:05

score 2 · Answer 2 · edited Mar 02 '20 at 16:25

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.

Square numbers multiplied by non-square numbers

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