goformit100
  • goformit100
Write whether the square of any positive integer can be of the form 3m+ 2, where m is a natural number. Justify your answer.
Mathematics
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katieb
  • katieb
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goformit100
  • goformit100
@genius12 Sir Please HELP me
mathslover
  • mathslover
Goformit, do you know euclid's division algorithm?
goformit100
  • goformit100
ya , but how to use it here ?

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mathslover
  • mathslover
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
  • goformit100
ok
anonymous
  • anonymous
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
  • mathslover
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
  • anonymous
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
  • mathslover
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
  • mathslover
@goformit100 , is your question correctly stated here? Re-check it.
mathslover
  • mathslover
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
  • Zarkon
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
  • mathslover
@Zarkon , Nice method. Thanks for introducing it to us.
mathslover
  • mathslover
@goformit100 , If the questions asks, that whether the given statement holds true for any positive integer, then NO! It is not correct.
mathslover
  • mathslover
But , @Zarkon , unfortunately, I am not good at congruent modulo. Can you teach me some of the basics?
Zarkon
  • Zarkon
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
  • goformit100
Thanks
mathslover
  • mathslover
Okay. I think, I will have to study it from any youtube lecture. Thanks @Zarkon ...
Zarkon
  • Zarkon
you could have also done cases on my last statement...what if m=0 or 1 or 2...sorta like you did
Zarkon
  • Zarkon
have you taken discrete math?
Zarkon
  • Zarkon
it is usually introduced in that class
mathslover
  • mathslover
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
  • mathslover
And Olympiad is also near, so I have to completely understand congruent modulo for solving tricky problems.
Zarkon
  • Zarkon
ic...good luck with that
mathslover
  • mathslover
Thanks :)

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