# Where does the constant increase by 2 of differences between integer square values come from?

$1^2 = 1$, $2^2 = 4$, $3^2 = 9$, $4^2 = 16$, $5^2 = 25$, etc...

Looking at the difference between those square values, we get: 3, 5, 7, 9, etc...

The difference from one (integer) square to the next increases by 2 without fault (let's assume).

Why is that? Why is there that pattern of increases by 2? What is it due to? What is the source of it?

I can "see" squaring visually as the construction of an actual square and I have drawn subdivided squares within squares to see the pattern unfold, but I just don't understand how to explain that increase by 2; I can't trace it, essentially.

• Hint. Can you calculate $(x+1)^2) - x^2$? May 7, 2015 at 16:36
• $(n+1)^2-n^2=2n+1$ May 7, 2015 at 16:36
• It's very much of the same origin as the second derivative of $f(x)=x^2$. May 7, 2015 at 16:55
• @JyrkiLahtonen, sorry I don't see it, do you mind explaining the connection? May 7, 2015 at 16:59
• Not him, but a derivative is infinitesimal difference whereas what you are looking at finite differences. May 7, 2015 at 22:12

For a visual intuition.... The inner blue region increments the gray square by one, but to make the next increment you need the extra yellow unit squares to get to the following square. • Thank you, this is exactly the kind of intuition I was looking for. May 7, 2015 at 18:15
• @jeremyradcliff There are many related illustrations on the web, see e.g. http://www.math.upenn.edu/~deturck/probsolv/LP1ans.html. May 8, 2015 at 14:39
• @JeppeStigNielsen, that's a great one too, thank you. May 8, 2015 at 16:10
• ... and if the increment is $k$ (instead of $1$), the second increment in squares is $2k^2$ bigger than the first. May 8, 2015 at 19:22
• IMHO even more intuition comes from shifting the yellow squares to the top right. May 12, 2015 at 3:49

We have the series $$x_n = n^2$$ and the differences $$\Delta x_n = x_{n+1} - x_n = (n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$$ and the difference of differences $$\Delta x_{n+1} - \Delta x_n = (2(n+1)+1) - (2n+1) = 2n + 3 - 2n - 1 = 2$$

The difference between two adjacent squares, $(a+1)^2$ and $a^2$:

\begin{align} (a+1)^2 - a^2 &= a^2 + 2a + 1 - a^2 \\ &= 2a + 1 \end{align}

So for each increase in $a$ by $1$, the delta between squares gets $2$ bigger.

Hint: $(a+1)^2-a^2=a^2+2a+1-a^2=2a+1$

Let $a$ be some natural number, and then let's think about $a+1$.
Since $a^2 = a \cdot a$, we have $(a+1)^2 = a^2 + 2a + 1$.

Then $(a+1)^2 - a^2 = 2a + 1$.

And then note for the numbers $1,2,3,\ldots$ applying $2a+1$ gives $2\cdot 1 + 1, 2 \cdot 2 + 1, 3 \cdot 2 + 1,\ldots$ Which is: $$3,5,7,9,\ldots$$