I know that I should use the definition of an odd integer ($2k+1$), but that's about it.
Thanks in advance!
I know that I should use the definition of an odd integer ($2k+1$), but that's about it.
Thanks in advance!
Step 1: pick an odd number (like $n=13$ here)
Step 2: bend it in "half" (any odd number $n$ can be written as $2k+1$, and $13=2\cdot 6 + 1$)
Step 3: fill in the blank space
Step 4: Count squares. (Here, the blue square has area $36=6^2$, while the whole square has area $49=7^2$)
Hint: Consider the difference of two consecutive squares. What is $(k+1)^2-k^2$?
HINT: $\begin{align} &2k + 1 \\= & 1\cdot(2k + 1) \\ =& \left(k + 1 - k \right)\left(k + 1 + k\right) \\ = & \cdots\end{align}$
Here is a more calculational answer where we try to 'construct' the solution by minimizing the amount of 'magic' in the proof. I'm assuming all variables are integers.
Given an odd $\;n\;$, we are asked to find $\;a,b\;$ such that $ a^2 - b^2 = n $
Let's calculate: \begin{align} & a^2 - b^2 = n \\ \equiv & \;\;\;\;\;\text{"arithmetic -- the simplest thing I know about the difference of squares"} \\ & (a+b) \times (a-b) = n \\ \equiv & \;\;\;\;\;\text{"arithmetic -- the simplest thing to give both sides similar structure"} \\ & (a+b) \times (a-b) = n \times 1 \\ \Leftarrow & \;\;\;\;\;\text{"logic: Leibniz -- the simplest thing to exploit the structural similarity"} \\ & a+b = n \;\land\; a-b = 1 \\ \equiv & \;\;\;\;\;\text{"arithmetic: solve for $\;a,b\;$"} \\ & a = (n+1)/2 \;\land\; b = (n-1)/2 \\ \end{align} Now since $\;n\;$ is odd, both $\;(n+1)/2\;$ and $\;(n-1)/2\;$ are integers, and therefore we have found the required $\;a,b\;$.
The algebraic equivalent of orlandpm's very elegant illustration is this:
Let a be any integer . . .
$K = (a+1)^2 - a^2$
$K = (a^2 + 2a + 1) - a^2$
$K = a^2 + 2a + 1 - a^2$
$K = 2a + 1$, which defines any odd integer; Q.E.D.
To get the complete classification of differences of Thai squares, we use the elementary observation that $a^2-b^2=(a+b)(a-b)$.
Clearly, $a+b$ and $a-b$ at either both even or both odd. So a d difference of two squares is either of our a multiple of $4$.
In fact, if $N$ is odd or a multiple of $4$, then we can always find $a$, $b$ such that $N=a^2-b^2$: if $N$ is odd, take $a=(N+1)/2$ and $b=(N-1)/2$, and if $N$is sa multiple of $4$, take $a=N/2+1$ and $b=N/2-1$.
Eric and orlandpm already showed how this works for consecutive squares, so this is just to show how you can arrive at that conclusion just using the equations.
So let the difference of two squares be $A^2-B^2$ and odd numbers be, as you mentioned, $2k+1$. This gives you $A^2-B^2=2k+1$.
Now you can add $B^2$ to both sides to get $A^2=B^2+2k+1$. Since $B$ and $k$ are both just constants, they could be equal, so assume $B=k$ to get $A^2=k^2+2k+1$. The second half of this equation is just $(k+1)^2$, so $A^2=(k+1)^2$, giving $A = ±(k+1)$, so for any odd number $2k+1$, $(k+1)^2-k^2=2k+1$.
You could add.
For mod ((2x+1),4)= 1, 2x+1 has at least one more solution than (k^2+1)-k^2=2x+1 and k^2 will be odd and (k-n)^2 will be even. Opposite if Mod((2x+1),4)= 3. n is always odd.
k^2-(k+n)^2= 2x+1, if 2x+1 is a composite number, if not, 2x+1 only has the solution (k^2+1)-k^2=2x+1 and 2x+1 is a prime number.
Sorry Mathjax is not working on my computer.