# Consecutive Square Numbers divided by Consecutive Odd Numbers

Why do we get this pattern (note what has been highlighted or italicized) when you divide consecutive square numbers with consecutive odd numbers ?

4 ÷ 3 = 1 R 1

9 ÷ 5 = 1 R 4

16 ÷ 7 = 2 R 2

25 ÷ 9 = 2 R 7

36 ÷ 11 = 3 R 3

49 ÷ 13 = 3 R 10

64 ÷ 15 = 4 R 4

81 ÷ 17 = 4 R 13

100 ÷ 19 = 5 R 5

121 ÷ 21 = 5 R 16

144 ÷ 23 = 6 R 6

169 ÷ 25 = 6 R 19

196 ÷ 27 = 7 R 7

225 ÷ 29 = 7 R 22

256 ÷ 31 = 8 R 8

Your observation (in the bold) is that $$(2n)^2$$ divided by $$4n-1$$ has the result $$n$$ R $$n$$.
$$(2n)^2=4n^2$$ and if you multiply the divisor $$(4n-1)$$ by the quotient $$n$$ and add the remainder $$n$$ you get $$((4n-1)\cdot n)+n=4n^2-n+n=4n^2$$.
Your observation (in italics) is that $$(2n+1)^2$$ divided by $$4n+1$$ has the result $$n$$ R $$(3n+1)$$.
$$(2n+1)^2=4n^2+4n+1$$ and if you multiply the divisor $$(4n+1)$$ by the quotient $$n$$ and add the remainder $$3n+1$$ you get $$((4n+1)\cdot n)+3n+1=4n^2+n+3n+1=4n^2+4n+1$$.