Odd numbers expressed as : $x^2-y^2$

Question

How to prove following statement :

Conjecture:

An odd number $n$ , $(n>1)$ can be uniquely expressed as : $n= x^2-y^2$ ; $x,y \in \mathbb{Z}^{*}$

if and only if $n$ is a prime number .

If $x-y=m$ , where $m>1$ then $m \mid n$

Proof :

$n=x^2-y^2=(y+m)^2-y^2=y^2+2\cdot y\cdot m +m^2-y^2 \Rightarrow$

$\Rightarrow n=m\cdot (2\cdot y+m) \Rightarrow m \mid n$

Therefore , if $m \neq 1$ it follows that $n$ is a composite number , but how to prove that every odd

composite number ,other than $1$ , has representation : $x^2-y^2$ , where $x-y>1$ ?

Answer 1

HINT $\ $ (Non) uniqueness follows from the composition law for differences of squares

$\rm\qquad\qquad\ (a^2-b^2)\ (A^2-B^2)\ =\ (a\:A+b\:B)^2-(a\:B+A\:b)^2$

$\rm\qquad\qquad\ \phantom{(a^2-b^2)\ (A^2-B^2)}\ =\ (a\:A-b\:B)^2-(a\:B-A\:b)^2$

E.g. composing $\rm\ 7 = 4^2 - 3^2\ $ with $\ 11 = 6^2 - 5^2\ $ yields for $\rm\: 7\cdot 11\:$ the following $\,2\,$ rep's

$\rm\qquad\qquad\ (4^2-3^2)\ (6^2-5^2)\ =\ (4\cdot 6+3\cdot 5)^2-(4\cdot 5+6\cdot 3)^2\ =\ 39^2 - 38^2$

$\rm\qquad\qquad\ \phantom{(4^2-3^2)\ (6^2-5^2)}\ =\ (4\cdot 6-3\cdot 5)^2-(4\cdot 5-6\cdot 3)^2\ =\ 9^2 - 2^2$

Remark $ $ Just like the e Brahmagupta–Fibonacci identity for composition of sums of squares can be viewed as arising from the multiplicativity of the norms of Gaussian integers, the above can be viewed as arising from norm multiplicativity of split-complex numbers (aka double or perplex numbers), $\,a+b\:\!j\,$ where $\,j^2 = 1.\ \ $

Answer 2

Write $n=ab$ with $b>1$ and $x^2-y^2=(x+y)(x-y)$. Solve $x+y=a$, $x-y=b$. This system has integer solutions $x=(a+b)/2$ and $y=(a-b)/2$ because $a$ and $b$ are odd and hence $a\pm b$ is even. Finally, $x-y=b>1$, as required.

Answer 3

If $n$ is odd and factors nontrivially as $ab$ with $a\le b$, then $n$ is the sum of $a$ consecutive odd numbers of which $b$ is the center one. Since the odd numbers are the first differences of the sequence of perfect squares, this means that $n$ is $(x+a)^2-x^2$ for some integer $x$.

Answer 4

We publish the theorem you are looking for here: WOLF Marc, WOLF François, Representation theorem of composite odd numbers indices, SCIREA Journal of Mathematics. Vol. 3 , No. 3 , 2018 , pp. 106 - 117 .

We proved that every odd composite number has representation : $x^2−y^2$ , where $x−y>1$

Answer 5

I tried to solve & got somewhere & I am presenting the same. This may be very long way, but it may be someway ...

We want prime numbers only by performing $x^2 - y^2$. Since all primes (except $2$) are odd, this suggests that both $x$ and $y$ are neither even nor odd. One is odd and other is even

So, Let us take $x = 2k_2 + 1$ and $y = 2k_1$, with $k_2 \geq k_1$, $k_1,k_2 \in \mathbb{N}$ (This is necessary since in $x^2 - y^2, y^2$ must not be $0$, so $y$ must not be $0$, so $k_1$ must not be $0$) (And this assumption does not take all odd numbers and so not all primes, but later we can reverse the role of $x$ and $y$)

Now, our final result $N = x^2 - y^2$ is a prime number. And we need to prove that for every prime number, we only have a unique $x$ and $y$.

Let $k_2 = k_1 + m, m \in \{0,1,2,\cdots\}$
Replacing $k_2$,
we get $N = 4m^2 + 8k_1m + 4k_1 + 4m + 1$ which is a prime number.

It is observed that the unique solution for all primes should have $x$ and $y$ as consecutive numbers. So if we prove that is $N$ has to prime, then $m$ has to $0$ for any $k_1$ that makes $N$ prime then we are done.

Taking $k_1$ as $k$,

$N = 4m(m + 2k + 1) + 4k + 1$

If we take $m = 0, N = 4k + 1$ which is superset of all primes (except $2$).

But then we need to prove that above $N$ can never be prime for any other value of $m$.