# Pattern for square numbers

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.

• 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 6 '18 at 15:27
• 11 has a red dot, but you skipped it entirely... – Gaurang Tandon Feb 6 '18 at 15:28
• No I have marked 11 in red only, 25+11=36... – naveen dankal Feb 6 '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... – naveen dankal Feb 6 '18 at 15:32
• $n^2 + (2n+1) = (n+1)^2$, 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 6 '18 at 15:34

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 $1764 = 42^2$.
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 $n^2$
$$\sum_{i=0}^n (2i+1)=2\sum_{i=0}^n i -\sum ^n 1$$ $$=2\frac {n(n+1)}{2} - n$$ $$=n^2+n-n= n^2$$