Write whether the square of any positive integer can be of the form 3m+ 2, where m
is a natural number. Justify your answer.

Question

Write whether the square of any positive integer can be of the form 3m+ 2, where m
is a natural number. Justify your answer.

goformit100 · Answer

@genius12 Sir Please HELP me

mathslover · Answer

Goformit, do you know euclid's division algorithm?

goformit100 · Answer

ya , but how to use it here ?

mathslover · Answer

It says, any positive integer a can be re-written as : $\bf{a=bq+r}$ (a is divided by b, having quotient q and remainder r ) . - Short definition for euclid's division algorithm.

goformit100 · Answer

ok

anonymous · Answer

If m is a natural number it belongs to the set N = {0, 1 , 2, 3...}
Let's try prove this statement through contradiction, i.e proof by contradiction.

To build this proof, we must contradict the statement, i.e. we must assume that 3m + 2 is not the square of any positive integer where m is a natural number.$$\bf 3m+2 \ne x^2, \ x \in \mathbb{Z}^+ $$Choosing a value for m, such as 5 gives us:$$\bf 17 \ne x^2, \ x \in \mathbb{Z}^+$$And this is true since 17 is not a perfect square.

@goformit100

mathslover · Answer

Now, say the number is a.

When, a is divided by 3, I can write it as:

$\mathsf{a=3q+r}$ 

Note that, remainder can not exceed 3 and will not be negative.
$\mathsf{0 \le r < 3}$ 

So, possible cases are :

$\mathsf{a =3q + 0\

a = 3q + 1\

a = 3q + 2}$

anonymous · Answer

So since the contradiction is correct, the original statement must be false.
Hence the square of any positive integer cannot be of the form 3m+ 2, where m
is a natural number.

mathslover · Answer

Square it now: 

Case 1 : a = 3q

$a^2 = 9q^2$ 
$a^2 = 3(3q^2) $ 

$q^2$ is also an integer.

Let us say that, $3q^2$ = m 

So, $a^2 = 3m$

mathslover · Answer

@goformit100 , is your question correctly stated here? Re-check it.

mathslover · Answer

Case : 2

$\mathsf{ a = 3q + 1\
a^2 = (3q+1)^2 \
a^2 = 9q^2 + 1 + 6q \
a^2 = 9q^2 + 6q + 1\
a^2 = 3(3q^2 + 2) + 1 \
a^2 = 3m + 1}$ 

Case : 3

$\mathsf{a = 3q + 2\
a^2 = (3q+2)^2\
a^2 = 9q^2 + 4 + 12q \
a^2 = 3(3q^2 + 4q + 1) + 1\
a^2 = 3m+1}$ 

Square of any positive integer can be in the form of 3m or 3m+1. But not 3m+2.

zarkon · Answer

a little quicker would be to ...

$$a=3q+m$$

$$a^2=9q^2+6mq+m^2$$

$$a^2=3(3q^2+2mq)+m^2$$

and then notice that $m^2\cancel{\equiv}2\text{  mod }3$

$$m=0,1,2$$

mathslover · Answer

@Zarkon , Nice method. Thanks for introducing it to us.

mathslover · Answer

@goformit100 , If the questions asks, that whether the given statement holds true for any positive integer, then NO! It is not correct.

mathslover · Answer

But , @Zarkon , unfortunately, I am not good at congruent modulo. Can you teach me some of the basics?

zarkon · Answer

i'll just write out what we have above

$$0^2\equiv 0 (mod 3)$$
$$1^2\equiv 1 (mod 3)$$
$$2^2\equiv 1 (mod 3)$$

goformit100 · Answer

Thanks

mathslover · Answer

Okay. I think, I will have to study it from any youtube lecture. Thanks @Zarkon ...

zarkon · Answer

you could have also done cases on my last statement...what if m=0 or 1 or 2...sorta like you did

zarkon · Answer

have you taken discrete math?

zarkon · Answer

it is usually introduced in that class

mathslover · Answer

Oh! I am in high school and we have to study all type of mathematics we have (not all but those who are of our level) ... So, no division, only study and study all topics.

mathslover · Answer

And Olympiad is also near, so I have to completely understand congruent modulo for solving tricky problems.

zarkon · Answer

ic...good luck with that

mathslover · Answer

Thanks :)