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I have found this on internet as a very interesting mathematical curiosity...

If we have $x\in \mathbb{N}$, they say that if $x=\sum^{K}_{n=0}{(2n+1)}$, then $\sqrt{x}\in \mathbb{N}$ and $\sqrt{x}=K+1$.

Is that always true? Why?

This is my first question here! Thanks in advance and sorry for any mistakes.

bertozzijr
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    You are absolutely correct! This has been discussed on this site many, many times before, and you will probably enjoy reading some of the many explanations that have been given – MJD May 28 '17 at 00:52
  • Nice, thanks for the heads-up! This is very good, indeed! – bertozzijr May 28 '17 at 00:55

3 Answers3

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Hint, look at the sequence of squares:

$0, 1, 4, 9, 16, 25$

Then look at the differences:

$1, 3, 5, 7, 9$

Formally, $(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$. Does that illuminate the problem?

orlp
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That is my favorite fact!

Consider this.

It's true that $1=1^2$

And it is true that $1+3=2^2$ and that $1+3+5=3^2$

Suppose it were true that $1+3+5+.....+(2n-1)=n^2$.

Then $1+3+5+.....+(2n-1)+(2n+1)=$

$n^2+(2n+1)= $

$n+2n+1=(n+1)^2$.

So if it's true for some it is true for the next.

Also.

Consider:

$N= 1+2+......+n $

Then $2N=N+N=$

$1+2+......+n+$

$n+......+2+1=$

$(n+1)+(n+1)+..... +(n+1) =$

$n (n+1) $

So $N=n (n+1)/2$

So $\sum (2k-1)=2\sum k -\sum 1$

$=2*\frac {n (n+1)}2-n$

$=n (n+1) - n=n^2+n-n=n^2$.

I love it.

fleablood
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Hint: the sum of an arithmetic progression of $N$ terms is $N$ times the average of the first and last terms. In this case: $\;x = (K+1) \cdot \cfrac{1 + (2K+1)}{2}\;$.

dxiv
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