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I know that I should use the definition of an odd integer ($2k+1$), but that's about it.

Thanks in advance!

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    Ohhhhhh I see, gotcha thanks for the tips yo!2013-01-16

8 Answers 8

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Step 1: pick an odd number (like $n=13$ here)

13 squares

Step 2: bend it in "half" (any odd number $n$ can be written as $2k+1$, and $13=2\cdot 6 + 1$)

enter image description here

Step 3: fill in the blank space

enter image description here

Step 4: Count squares. (Here, the blue square has area $36=6^2$, while the whole square has area $49=7^2$)

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    @orlandpm how did you do the illustration? i love it!2013-08-26
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Hint: Consider the difference of two consecutive squares. What is $(k+1)^2-k^2$?

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    @papercuts Try it with the difference of _any_ two squares: $(k+n)^2-k^2$. Then let n=1.2012-12-21
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HINT: $\begin{align} &2k + 1 \\= & 1\cdot(2k + 1) \\ =& \left(k + 1 - k \right)\left(k + 1 + k\right) \\ = & \cdots\end{align}$

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    nicely proved !2014-06-20
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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 $\;a,b\;$.

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    You seem to be confusing the arrow directions. This step says that _if_ $\;a+b=n\;$ and $\;a-b=1\;$, _then_ also the product of the left hand sides equals the product of the right hand sides. Leibniz' rule just says in general that $\;x = y\;$ implies $\;\ldots x \ldots \;=\; \ldots y \ldots\;$. The philosophical name is "[indiscernibility of identicals](https://en.wikipedia.org/wiki/Identity_of_indiscernibles)".2016-01-02
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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.

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    Eric had already hinted that. And bringing old posts up by answering them without bringing something really new isn't something you should do. Also, this site uses latex math-mode to format maths. (Just put `$`s around your formulas)2013-07-27
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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 $b=N/2-1$.

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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 $2k+1$, $(k+1)^2-k^2=2k+1$.

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    @msh210 Neither of them are "fixed" yet; if they were, they would have set values and wouldn't be very usable for proofs (and, barring certain specific values, would probably be referred to with that value instead). I edited my answer to try and clarify (since it's more of an assumption than a known).2012-12-21
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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.

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    Who are $x$ and $n$? The problem speaks only about $k$. What do you mean by "solution", given that there are no equations involved either in the question, or in your own answer?2015-11-20