# Find all even numbers that can be represented as a difference of squares in only two ways

I am currently working on this proof. I am looking to find (with proof) all even numbers that can be represented as a difference of squares in only two ways.

My thoughts thus far.

I examined the first 40 natural numbers and found that 16, 24, 32, 40 have exactly two representations as a difference of squares. My claim is that all numbers $x$ for which $x=8k+16$ will have exactly two representations.

$8k+16=(k+a)^2 - (k+b)^2$ $\rightarrow$ $8k+16=(k+4)^2 - k^2$ which is the first representation.

My idea for finding the second representation is to express $8k+16$ as

$8k+16=(ak+b)^2 - (ck+d)^2$

Expanding and attempting to solve I get:

$$\left.\begin{matrix} a^2 - c^2 = 0\\ 2ab - 2cd= 8\\ b^2 - d^2= 16 \end{matrix}\right\}$$

This is where my problem arises, $b^2-c^2=(b-c)(b+c)=16=4^2$, I cannot seem to move on from here.

If anyone can offer a hint that would help with my understanding of the problem and the attempt at the proof, that would be extremely helpful. Thank you.

• Note that $36=6^2-0^2=10^2-8^2$. – André Nicolas Jul 14 '16 at 19:40

We have $$2n=a^2-b^2=(a-b)(a+b)$$ $a-b$, $a+b$ must be of the same parity, and that parity must be even, since otherwise their product wouldn't be the even number $2n$. Hence $n$ is even, so we may as well have $$4m=(a-b)(a+b)$$ or $$m=\frac{a-b}{2}\frac{a+b}{2}$$

Hence, we factor $m=st$, with $s\le t$, and set $\frac{a-b}{2}=s$, $\frac{a+b}{2}=t$, or $a=s+t$, $b=t-s$.

For $16$, we have $4=1\cdot 4=2\cdot 2$. This leads to $\{a=5, b=3\}$ and $\{a=4,b=0\}$.

For $24$, we have $6=1\cdot 6=2\cdot 3$.

For $32$, we have $8=1\cdot 8=2\cdot 4$.

It appears that the desired $n$ are of the form $4pq$, for $p,q$ not necessarily distinct primes, and of the form $4p^3$. $pq$ has factorizations $1\cdot pq=p\cdot q$, and $p^3$ has factorizations $1\cdot p^3=p\cdot p^2$. It's a bit tedious to look at various cases, but I'm fairly sure that all other forms for $n$ will have more than two factorizations, leading to more than two $\{a,b\}$ pairs. Also not proved (but I think not too hard) is that two different factorizations must lead to different $\{a,b\}$ pairs.

• Amusing, we used the same letters, $m$, $s$, $t$. A uniform culture! – André Nicolas Jul 14 '16 at 20:02
• Could another form be $4p^2$ since the only factorizations are $1 \cdot p^2$ and $p \cdot p$. – Joe Jul 14 '16 at 20:52
• @Joseph that's covered by the $4pq$ case; it is even made explicit that the primes $p,q$ need not be distinct. – quid Jul 14 '16 at 22:17

Outline: As you know, we are solving $(a-b)(a+b)=n$, with the side requirement that $a+b$ (and therefore $a-b$), are even.

So $n$ is of the shape $4m$. If $st=m$, we get the solution $a-b=2s$, $a+b=2t$. Thus the number of representations is the number of ways to express $m$ as a product $st$ with $0\lt s\le t$.

Now there is some work for you to do. What numbers $m$ can be expressed in precisely two ways as $st$, where $0\lt s\le t$?

• @Henry: Thanks, need to interchange. – André Nicolas Jul 15 '16 at 16:20

My claim is that all numbers $x$ for which $x=8k+16$ will have exactly two representations.

Counterexample: \begin{align} 120 &= 31^2-29^2 \\ &= 17^2 - 13^2 \\ &= 13^2-7^2 \\ &= 11^2-1^2 \end{align}

Note that while I just picked a number divisible by $8$ with plenty of factors, the claim actually falls at $48$: \begin{align} 48 &= 13^2-11^2 \tag{from (12,1)}\\ &= 8^2 - 4^2 \tag{from (6,2)}\\ &= 7^2-1^2 \tag{from (4,3)}\\ \end{align} because $48/4=12$ has three pairs of factors.

As Joffan pointed out, your conjecture isn't true. Here's my attempt at finding the correct condition for a positive even integer $n$ to be represented by a difference of squares in exactly 2 ways.

Suppose $n= a^2-b^2=(a+b)(a-b)$. This suggests that factorizations of $n$ can be linked to representations as a difference of squares. Clearly each representation of $n$ as a difference of squares gives us a factorization of $n$. Which factorizations of $n$ give us a difference of squares?

Suppose $n=xy, x<y$, then we want to find $a,b$ such that $a+b=x,a-b=y$. We get $a=\frac{x+y}{2},b=\frac{x-y}{2}$. Since you said n is even, we know that $x,y$ must both be even.

Therefore, every factorization of $n$ such that each factor is even gives us a representation of $n$ as a difference of squares. So how many such factorizations does $n$ have?

Firstly, note that this is equal to the number of factorizations (without any restrictions on the factors) of $n/4$. The number of factorizations of any number is the number of factors that number has divided by two, rounded up (only needed in the case of squares).

There is a nice formula for the number of factors of $n/4$ in terms of its prime factorization. If $n/4=\Pi_{i=1}^n p_i^{e_i}$ then there are $\Pi_{i=1}^n (e_i+1)$ factorizations

We want 2 representations, and so we want 3 or 4 factorizations. With some casework, this gives: $n=4p^2,4p^3,4pq$ for $p,q$ prime.

• Is it sufficient to say that we can conclude that $n=4p^k$ for $k \geq 4$ will not have exactly two factorizations since $p^k$ for $k \geq 4$ can always be written as $p^2p^{k-2}$. Moreover, for $p^k$ for $k \in \mathbb{N}$, each $p^k$ will always have factorizations $1 \cdot p^k$ and $p^{k-1}p^k$. Thus $n$ = $4pq$, $4p^2$, $4p^3$ are the only even numbers $n$ that have exactly two factorizations and hence have two representations. – Joe Jul 14 '16 at 21:08

The first counterexample for which $8k+16$ is not uniquely described by a difference in squares in only two ways is $48$.

\begin{align}48&=8\cdot4+16\\&=7^2-1^2\\&=8^2-4^2\\&=13^2-11^2\end{align}

All product $(a+b)(a-b)$ is a such a difference $a^2-b^2$ so if your claim were true, the infinity of diophantine equations $$(x+y)(x-y)= 8k+16\text{ where }k\ge 1$$ would always have just two solutions; it is doubtful, in particular for $k$ even with many prime factors. Taking for example $k=2n$ one has $$(x+y)(x-y)=16(n+1)$$ Choose now $n+1=abc$ with $a,b,c$ even so you have the equation $$(x+y)(x-y)=16abc$$ This way you have several couples of even factors $A,B$ with $AB=16abc$ and for each of them the system $$\begin{cases}x+y=A\\x-y=B\end{cases}$$ gives a solution $(x,y)$ for which you have $$x^2-y^2=16abc$$ (don't forget, of course that our $16abc$ is by construction, of the form $8k+16$).

As others have pointed out, the conjecture is not true. I really enjoyed reading how Euler investigated the topic:

Euler Archive

E228 - On the numbers which are the sum of two squares (1758)

E134 - Theorems on the divisors of numbers (1750)

Subject, Number Theory, English