# Prove every odd integer is the difference of two squares

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

• Hi, papercuts. You might want to start thinking about upvoting answers that are helpful (click on the upward arrow (grey) to the left of the answer. You can also accept one answer, which you can do by clicking on the "greyed-out" check-mark to the left of the answer you want to accept. upvoting and accepting answers encourages people to take the time needed to answer your questions, and is a way to show appreciation. – Namaste Jan 16 '13 at 2:59
• Ohhhhhh I see, gotcha thanks for the tips yo! – papercuts Jan 16 '13 at 3:01

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 $49=7^2$)

• +1 nice straight lines and corners, no wobbliness or smudges. – Graham Borland Dec 21 '12 at 13:53
• It's great to see a visual presentation. I never thought about it this way. – krikara Dec 21 '12 at 22:32
• Note $\$ One can find many analogous visual proofs in Conway and Guy's The Book of Numbers. Highly recommended for budding mathematicians. – Math Gems Feb 13 '13 at 18:42
• @Eric, I seem to recall an answer of yours that also does this... :) – J. M. is a poor mathematician May 18 '13 at 18:47
• @orlandpm how did you do the illustration? i love it! – zerosofthezeta Aug 26 '13 at 18:43

Hint: Consider the difference of two consecutive squares. What is $(k+1)^2-k^2$?

• Oh, haha well that was easy, but how'd you know to use consecutive squares? – papercuts Dec 21 '12 at 7:57
• @papercuts Probably mostly experience. There's not much substitute for it. – Alex Becker Dec 21 '12 at 8:00
• It's also one of the simplest sorts of differences of two squares to analyze. – user14972 Dec 21 '12 at 9:35
• @papercuts Try it with the difference of any two squares: $(k+n)^2-k^2$. Then let n=1. – Griffin Dec 21 '12 at 18:10

HINT: \begin{align} &2k + 1 \\= & 1\cdot(2k + 1) \\ =& \left(k + 1 - k \right)\left(k + 1 + k\right) \\ = & \cdots\end{align}

• Would the downvoter care to explain? – Parth Kohli May 18 '13 at 14:32
• nicely proved ! – Debashish Jun 20 '14 at 8:34

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 $\;a,b\;$.

• Could we please explain the Leibnitz step? I'm not sure how a + b = n and a - b = 1 follows from (a+b)×(a−b)=n×1. – mcandre Jan 2 '16 at 2:12
• There are plenty of operators (e.g. modulo) where f(a, b) = f(c, d) has solutions where (c, d) != (a, b). Is multiplication special this way? Would a + b = 1 and a - b = n also have integer solutions, since n*1 = 1*n? – mcandre Jan 2 '16 at 2:28
• 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". – MarnixKlooster ReinstateMonica Jan 2 '16 at 8:57

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.

• 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 $s around your formulas) – xavierm02 Jul 27 '13 at 23:59 To get the complete classification of differences of Thai squares, we use the elementary observation that$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$2k+1$,$(k+1)^2-k^2=2k+1$. • Re "Since B and k are both just constants, they can be treated as equal": no: you already fixed both. – msh210 Dec 21 '12 at 15:30 • @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). – MyNameIsNotMcThomasJohannson Dec 21 '12 at 15:37 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. • 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? – Alex M. Nov 20 '15 at 10:52