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I understand this is intuitively obvious. I am wondering if there is a axiom or principle or proof that shows this? It seems you may need something that says you can decompose an integer into it's lowest common factors and if a number is odd then each of it's lowest common factors will never be divisible by two which is the lowest possible even integer.

It seems like there may be something fundamental beside my intuition here in number theory maybe...and that I either forgot or never learned.

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    $\begingroup$ Use Euclid's division lemma and the definition of odd numbers that they are not divisible by 2 and proceed forward. $\endgroup$ – Abhinav Dhawan Nov 4 '17 at 16:13
  • $\begingroup$ I am ok with that....but will look at the answers below as well. thanx $\endgroup$ – Sedumjoy Nov 4 '17 at 18:37
  • $\begingroup$ there are many beautiful ways to do this and I feel stupid for not thinking of a way. $\endgroup$ – Sedumjoy Nov 4 '17 at 18:41
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    $\begingroup$ @Sedumjoy We all couldn't solve this problem at some point. Don't despair. $\endgroup$ – Duncan Ramage Nov 4 '17 at 19:22
  • $\begingroup$ Possible duplicate of Why are the last two digits of a perfect square never both odd? $\endgroup$ – NewBee Nov 4 '17 at 21:30
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Let $n$ be an odd integer. Then, there exists an integer $k$ such that $n = 2k + 1$. Thus, $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$. Since there exists an integer $l$ such that $n^2 = 2l + 1$, namely, $l = 2k^2 + 2k$, $n^2$ is odd.

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  • $\begingroup$ I lose you at the first condition in the last sentence. "n" is given as odd, whether n squared is odd or not is the question. "Since..." asserts n squared is odd, it doesn't prove it. (2 * any integer) + 1 is odd. (0+0+1, 1+1+1, 2+2+1, -1+-1+1, all odd, because any multiple of 2 is even, and +1 makes any even number odd.) "...namely," and what follows seem like arm waving. $\endgroup$ – Bill IV Nov 5 '17 at 0:25
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    $\begingroup$ @Bill The last sentence is an example of existential introduction. From the statement "$n^2 = 2(2k^2 + 2k) + 1$", we derive the statement "there exists $l$ such that $n^2 = 2l + 1$". $\endgroup$ – Tanner Swett Nov 5 '17 at 1:40
  • $\begingroup$ This answer would be better as a picture. $\endgroup$ – Solomonoff's Secret Nov 5 '17 at 3:00
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    $\begingroup$ @BillIV It isn't. $p$ is odd if and only if there exists an integer $q$ such that $p = 2q + 1$. I have shown there exists an integer $q$ such that $n^2 = 2q + 1$, so $n^2$ is odd. $\endgroup$ – Duncan Ramage Nov 5 '17 at 5:00
  • $\begingroup$ Duncan, it's the ordering. For the last sentence, I'd have said, "Since "k" is an integer, "((2k)^^2 + 2k)" is also an integer. Substitute integer "l" for "((2k)^^2) + 2k)", n2 = 2l + 1. n2 is odd. Note, your reply substitutes q for l. I also have trouble with the famous Pythagorean theorem proof that's 4 right triangles forming a square and enclosing a rectangle, at an angle. The title is "Behold!". I can work it out with effort, but the 3-4-5 triangle with squares of 9, 16 and 25 unit squares attached to their respective edges is something I can actually see and follow. $\endgroup$ – Bill IV Nov 7 '17 at 2:25
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An odd number can be represented as $2k{+}1$ with $k\in \Bbb Z$ (and an even number cannot). So squaring that expression $(2k+1)^2=4k^2+4k+1 = 2(2k^2+2k) + 1$ and clearly $(2k^2+2k)\in \Bbb Z$ so the square is odd.

Equivalently, you could work $\bmod 2$ since any odd number $n\equiv 1 \bmod 2$ and thus $n^2\equiv 1^2\equiv 1 \bmod 2 $ is odd also.

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Every number has a unique prime factorization (fundamental theorem of arithmetic). This means any odd number, $a$ is

$a = f_1^{p_1} * f_2^{p_2} * ... * f_N^{p_N}$

where none of the $f_i = 2$.

You can write $a^2$ in terms of the $f_i$ and see that

$a^2 = f_1^{2p_1} * f_2^{2p_2} * ... * f_N^{2p_N}$,

and again no prime factors are 2.

I.E. squaring does not change the spectrum of prime factors, only their powers, so it can't take a number from odd to even.

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1) The product of 2 odd numbers is odd.

$a=2k+1; b= = 2l+1.$

$ab= (2k+1)(2l+1) =$

$ 4kl + 2(k+l) +1 =$

$ 2[2kl +(k+l)] +1$, odd.

In your case $a=b$, hence $a^2$ is odd.

2) The product of 2 even numbers is even.

$a=2k; b=2l;$

$ab= 2(2kl)$, even.

3) The product of an even and an odd number is even.

$a=2k; b=2l+1;$

$ab= 2k(2l+1)= 2[k(2l+1)],$ even.

4) How about the product of an odd and an even number? :)

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It is possible to use the definition of a prime here - assuming you are allowed to know that $2$ is prime.

So $p$ is a prime if $p|mn$ implies that $p|m$ or $p|n$.

(In some approaches to arithmetic this is a theorem rather than a definition - or rather the theorem is that over the integers every irreducible is prime. Sometimes the properties of an irreducible are used as the definition of integer primes. It is only later that the distinction starts to become important.)

Apply this to $2|n^2$ to conclude that $2|n$ or $2|n$ i.e. $n$ is even.

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Observe that $$ m^2=2\binom{m+1}{2}-\binom{m}{1}=2\left(\frac{m(m+1)}{2}\right)-m. $$ When $m$ is odd $m^2$ thus equals an even number minus an odd number which is odd.

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    $\begingroup$ I feel stupid... there are so many elegant ways to show this and I did not think of them. ..I will have to study them all $\endgroup$ – Sedumjoy Nov 4 '17 at 18:40
  • $\begingroup$ This is by far the most clear and readable to someone with a physics background. I hope a more mathy person weighs in on whether it's "math valid". $\endgroup$ – user1717828 Nov 4 '17 at 19:53
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    $\begingroup$ It's valid if you prove a Lemma that the difference of an even number and an odd number is odd. $\endgroup$ – James S. Cook Nov 4 '17 at 21:26
  • $\begingroup$ Mmmm... this one is nice. I like it. $\endgroup$ – user1628658 Nov 9 '17 at 22:27
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Do you know that you can represent an odd integer using the following form: $2k + 1$ where $k=0,1,2,...$. In other words, $2k + 1$ in effect says that an odd integer is nothing more than an even integer plus one. So, what this all means is that if you pick any integer $k$ whatsoever from the set $\{0,1,2,...\}$, plug it in into the formula, as a result you get an odd number.

What if we now square it? Well, we get the following: $(2k + 1)^2=(2k)^2+2\cdot 2k \cdot 1 + 1^2=$ $4k^2+4k+1$. Now, placing everything else except the $1$ into a set of parentheses and factoring out a $2$, we get: $2(2k^2+2k)+1$. What you now observe is the fact that $2(2k^2+2k)+1$ is nothing but an even integer plus one which means that it's odd. This in turn means that the expression $(2k + 1)^2$ (the square root of an odd integer) we started with was an odd number to begin with. And that's how you prove that if an integer is odd, its square is an odd number too.

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Hm... I'm not sure whether I'll be able to write this one as I have it in my mind. Let's think of this as the answer for dummies:

a*b = a+a+a+a+a+...+a where we have b amount of a.

a^2 = a * a = a+a+a+a+a+...+a where we have a amount of a.

Let's consider a=(2k+1) - therefore an odd number, where k is of {0,1,2,3,...,inf}

Therefore we can write

a * a == (2k+1) * (2k+1) == 2k(2k+1)+ 1*(2k+1) == 4(k^2)+2k + (2k + 1) == 4(k^2) + 2k + 2k +1

Here we can by now clearly see, that whatever number k is, each has either a 4 or a 2 next to each other making them even because even*odd = even [ 1+1 = 2*1 = 2 ]

Note: I just realized that Peter Szilas wrote it a lot shorter than I did... I didn't actually rip off his answer. It occurred to me at this point of writing, that someone most probably thought of this as it's fairly elementary.

Back to the problem:

4(k^2) + 2k + 2k +1 == even + even + even + odd

Since there is only a single odd number in this sum, the result will be odd.

I'm aware that there is a more sophisticated way of explaining this, but It's been a long time since I used Math and English at the same time.

Edit: Actually this is a sub-problem of if a is odd b is odd then a * b = odd

... and someone else explained this as well as I see. Never mind.

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