1

enter image description here

Due apologies for this rustic image. But while drawing this lattice arrangement about the "square numbers" , I discovered a pattern here wherein if I add the alternate red dots (as depicted in the image above) to the square number, I get the next square number. For instance, $4 + 5(red\ dot) = 9$ , $9+7(red\ dot)=16$, $16+9(red\ dot)=25$, $25+11(red\ dot)=36$, $36+13 (red\ dot)=49$.

The red dotted numbers themselves have a pattern as is obvious from the image. Is there any mathematical explanation to this pattern.

  • 1
    How do you decide what is red and what is blue? What do you mean by alternate red dot? Eg 9,5 are red but 9+5=14 is not square. – Dan Robertson Feb 06 '18 at 15:27
  • 11 has a red dot, but you skipped it entirely... – Gaurang Tandon Feb 06 '18 at 15:28
  • No I have marked 11 in red only, 25+11=36... –  Feb 06 '18 at 15:29
  • @Dan Robertson my idea is that if i take these red dots in sequence and the square number in sequence too and sort of take a bijection, say align first square number, i.e. 4 with the next red dot , i.e., 5 I get the next square number. then take 9 and the next red dot after 5, i.e., 7 and add i get 16 and so on... –  Feb 06 '18 at 15:32
  • There are many versions of your question on this site. Essentially, you are adding the next odd number to each square number: https://math.stackexchange.com/questions/136237/direct-proof-that-1-3-5-cdots-2n-1-n-cdot-n – Ethan Bolker Feb 06 '18 at 15:33
  • 1
    $n^2 + (2n+1) = (n+1)^2$, i.e. to get from $n^2$ to the next square you have to add $2n+1$ which is the $n$th odd number. – Jaap Scherphuis Feb 06 '18 at 15:34

2 Answers2

0

It's a little hard to tell from the picture, but it seems like you've arranged the red dots so that all odd numbers get red dots, and all even numbers get blue. (It looks to me like 3 gets blue and 16 gets red, though, but they also don't fit the obvious pattern of the dots, so either I'm misinterpreting the pattern or they are mistakes.)

The $n$th square number is $n^2$, and the $n+1$st odd number is $2n+1$. If we line them up in such a way, then adding the $n+1$st odd number to the $n$th square number gives $n^2 + 2n +1 = (n+1)^2$, the $n+1$st square number.

Incidentally, this is a very efficient way of mentally calculating some square numbers. E.g., suppose you want to mentally calculate $42^2$. You start with $40^2$ which you know is $1600$, and then you add the $41$st odd nubmer, which is $40+41$. That gets you to $1681 = 41^2$. Now you add the $42$nd odd number, which is $41+42$.That gets you to $1764 = 42^2$.

BallBoy
  • 14,472
0

What you are doing is nothing but adding consecutive odd numbers to a perfect square which will obviously turn out to be a perfect square. I guess you might know that sum of first $n$ consecutive odd numbers is $n^2$

If not it can be proved easily like

$$\sum_{i=0}^n (2i+1)=2\sum_{i=0}^n i -\sum ^n 1$$ $$ =2\frac {n(n+1)}{2} - n$$ $$=n^2+n-n= n^2$$

Rohan Shinde
  • 9,737