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My question is simple: If all integers can be represented by $2k$ or $(2k+1)$ where all odd numbers have the form $(2k+1)$ the square of any odd number should be in the form of $(4k^2 + 4k +1) = 4q+1$ then why we have to further represent $q$ (in the last equation) as $2k$ and $2k+1$ to make a conclusion that odd number squared is always is in the form of $8q+1$ ? didn't we already assumed that any odd number is in the form $2k+1$ where $k$ can be any integer. if that isn't the case then why should not we further replace $q$ as $2k$ and $2k+1$ and further and this wont stop. so what is logic here?

The proof makes sense if we start with odd integer as $4k+1$ or $4k+3$ but again how we decide which form should we start with, because both (2k and 4k version) represent the odd numbers.

Bill Dubuque
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Asim
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    Who says you bave to check the parity of $q$? In what context? We don't know what you are looking at so perhaps they are after some particular thing you didn't mention? – lulu Apr 17 '21 at 10:53
  • i don't understand what u asking, but this is from the elementary number theory book. its asks to show that square of any odd number is always in the form of 8q+1 – Asim Apr 17 '21 at 10:57
  • Yes the square of an odd number is of the form 8q+1 where $q$ is an integer number (which may be either odd or even). – user Apr 17 '21 at 10:58
  • The title makes no sense since the square of an odd number IS of the form $4k+1$ , in fact of the form $8k+1$ which is extremely easy to see : $(2x+1)^2=4x^2+4x+1=4x(x+1)+1$. So , you only have to show that $x(x+1)$ must be even. Can barely be easier. – Peter Apr 17 '21 at 11:00
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    Well, if you are trying to prove that it is $8k+1$, then it is not enough to show that it is $4q+1$. Right? – lulu Apr 17 '21 at 11:04
  • @Peter i am new to number theory, if we assume k equals 1 than it would be 5 and contradicts the equation, but from above answers it seems like we have to rely on our intuition rather than logical proof to know which form it should be. – Asim Apr 17 '21 at 11:10
  • @Asim Could you please edit the question? It's not making sense – BooleanCoder Apr 17 '21 at 11:11
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    It is easy to see that the square of an odd number is odd. But can we say a little more? Yes! By your argument, we can see that the square of an odd number is of the form $4q+1$. Does that make the first statement false? Not at all, but the new statement is stronger. Can we say even more, then? Yes! By the slightly improved version of your argument, we can show that the square of an odd number has form $8q+1$. Stronger (but does not make the previous results wrong). Can we then go even further and show that squares of odd numbers have form $16q+1$. No! Easy to find counterexamples. – Jeppe Stig Nielsen Apr 17 '21 at 11:12
  • Not every number of the form $4k+1$ is an (odd) perfect square. This direction is in fact wrong. But what you want to prove is the other direction : Every odd perfect square is of the form $4k+1$ ( which can be extended to $8k+1$ ) – Peter Apr 17 '21 at 11:16
  • @JeppeStigNielsen i know you are right but, the confusion is, do we always make this trial and error method to reach the final conclusion or is there any solid formula to know when to stop. do we always do it with odd number representation or with even numbers too? one other similar long example shows the final form in 8q+1 when i did the calculation but the solution in the book expands it further and says it should be 16q+1 . – Asim Apr 17 '21 at 11:23
  • Well, if you want to work $\pmod 8$, then you can just work $\pmod 8$. Every odd number is of one of the forms $8k+1,8k+3,8k-3,8k-1$. You could just check each of those forms to see that they all square to something of the form $8k+1$. As it happens, it suffices to work $\pmod 4$, but you could just work $\pmod 8$ if you preferred. – lulu Apr 17 '21 at 11:26
  • More broadly, though, your question is not clear. Proofs aren't always easy. Sometimes your first idea doesn't work. That's just the way it is. – lulu Apr 17 '21 at 11:27

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It is of that form:

$$(2k+1)^2=4(k^2+k)+1=4q+1\;,\;\;\text{with}\;\;q=k^2+k\in\Bbb N$$

And since $\;q=k^2+k=k(k+1)\;$ is always even, the above in fact can be written as

$$k^2+k=2m\implies (2k+1)^2=4(k^2+k)+1=8m+1$$

and you get at once that any odd number squares equals $\;1\pmod 8\;$ .

DonAntonio
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  • this makes me confuse because many other examples doesn't care about what q is equal to. it just concludes that the expression is a multiple of k times some q without further knowing what is q. i hope i made my point clear. – Asim Apr 17 '21 at 11:04
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    In order to write your odd number squared as $;4q+1;$ nobody cares what $;q;$ is except for the fact that it must be a natural number. If you want to show your odd number squared equals $;8m+1;$ , with $;m\in\Bbb N;$ , then one must show that $;q;$ above is even. That's all... – DonAntonio Apr 17 '21 at 11:08
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    @Asim It depends entirely on what you want to do with the odd square later. Many times it's enough to know it's of the form $4q+1$, some times we need $8m+1$. If we only need $4q+1$, then going all the way to $8m+1$ is likely to be confusing. – Arthur Apr 17 '21 at 11:21
  • @Asim Does my answer, answer your question? – BooleanCoder Apr 17 '21 at 11:30
  • @DonAntonio because my given example is simple it is even more obvious that if we take 0 or 1 as the k (in above equation) the whole expression becomes 8k+1, but problem is if the question was very complicated how can we decide which form is more stronger in proof. i hope u understand my point. and thank you all for the answers. – Asim Apr 17 '21 at 11:48
  • @Asim I honestly don't understand what you're asking: what "given example" are you talking about? You usually cannot get to choose $;k;$ : it is what it is that's all there is. For example, with $;k=6;$ we'de get $$13^2=(2\cdot 6+1)^2=4(6^2+6)+1=8\cdot21+1$$ and you choose what expression to use. – DonAntonio Apr 17 '21 at 12:21
  • @DonAntonio what i meant was in the expression $4k^2+4k+1$ if we take the minimum value of k like 0 or 1 than this expression has the minimum value as $8k+1$ for any k so this can't be lowest than that and we can take multiple of that expression to any k and get the same $8k+1$ form, – Asim Apr 17 '21 at 12:30
  • @DonAntonio also you said " if we want prove odd integer squared has the 8k+1" in that case we are already trying to get 8k+1 and looks easier, but my question is in general: what if we want to know what the actual form an odd integer squared (or to any power) will have? than how to decide to either start with odd integer as 2k+1 or as 4k+1 and 4k+3 .. – Asim Apr 17 '21 at 12:42
  • @Asim All this is very, very confusing. You wrote "what i meant was in the expression 4k2+4k+1 if we take the minimum value of k like 0 or 1 than..." ...once again, if you're given an odd integer then you've no choice how you write it as $;2k+1;$...it is GIVEN ! I can't really grasp what you say about minimum value (?!?) and all the rest.... – DonAntonio Apr 17 '21 at 12:45
  • @DonAntonio from the given example we reach at the point where it says an odd integer squared has the form $4k^2+4k+1$ obviously k can by any integer from 0 to infinity and the equation holds true. if for example we choose k as 0 then result would be $(4(0)^2 +4(0)+1) = 1$ or k as 1 then $(4(1^2) + 4(1) +1)= 8+1 $ or k as 2 then $ 4(4)+4(2)+1 = 25 $ this is same as 8q+1 . – Asim Apr 17 '21 at 13:11
  • @Asim There is no "given example" at all, and yes: depending on what the odd number is, that $;k;$ can be any integer ...so what?? – DonAntonio Apr 17 '21 at 13:17