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Nyblom, M. A. On the representation of the integers as a difference of nonconsecutive triangular numbers. Fibonacci Quarterly 39 (2001), no. 3, 256–263.
Theorem 2.1: Let $M \in \mathbb{Z} \backslash\{0\}$, then the number of distinct representations of $M$ as a difference of nonconsecutive triangular numbers is given by $N_{\Delta}(M)=D-1$, where $D$ is the number of odd divisors of $M$.
Proof: Without loss of generality, we may assume that $M$ is a positive integer. Our aim here will be to determine whether there exists $x, y \in \mathbf{N} \backslash\{0\}$ such that $M=T(x)-T(y)$. By completing the square, observe that the previous equation can be recast in the form $8 M=X^{2}-Y^{2}$, where $X=2 x+1$ and $Y=2 y+1$. To analyze the solvability of this equation, suppose $a b=8 M$, where $a, b \in \mathbf{N} \backslash\{0\}$ and consider the following system of simultaneous linear equations:
$$\tag1\label1
\begin{aligned}
&X-Y=a \\
&X+Y=b
\end{aligned}
$$
whose general solution is given by
$$
(X, Y)=\left(\frac{a+b}{2}, \frac{b-a}{2}\right) .
$$
Now, for there to exist a representation of $M$ as a difference of nonconsecutive triangular numbers, one must be able to find factorizations $a b=8 M$ for which the system \eqref{1} will yield a solution $(X, Y)$ in odd integers.
Remark 2.1: We note that it is sufficient to consider only \eqref{1}, since if for a chosen factorization $a b=8 M$ an odd solution pair $(X, Y)$ is found, then the corresponding representation $M=T(x)-$ $T(y)$ is also obtained if the right-hand side of \eqref{1} is interchanged. Indeed, one finds upon solving
$$
\begin{aligned}
&X^{\prime}-Y^{\prime}=b \\
&X^{\prime}+Y^{\prime}=a
\end{aligned}
$$
where $X^{\prime}=2 x^{\prime}+1, Y^{\prime}=2 y^{\prime}+1$ that
$$
X^{\prime}=\frac{a+b}{2} \text { and } Y^{\prime}=\frac{a-b}{2}=-Y .
$$
Thus, $x^{\prime}=x$ while $y^{\prime}=(-Y-1) / 2=-y-1$, so
$$
T\left(y^{\prime}\right)=\frac{(-y-1)(-y)}{2}=T(y) \text { and } T\left(x^{\prime}\right)-T\left(y^{\prime}\right)=T(x)-T(y)=M .
$$
We deal with the existence or otherwise of those factorizations $a b=8 M$ which give rise to an odd solution pair $(X, Y)$ of \eqref{1}. It is clear from the general solution of \eqref{1} that, for $X$ to be an odd positive integer $a, b$ must be at least chosen so that $a+b=2(2 s+1)$ for some $s \in \mathbb{N} \backslash\{0\}$. As $a b$ is even, this can only be achieved if $a$ and $b$ are also both even. Furthermore, such a choice of $a$ and $b$ will also ensure that $Y=X-\alpha$ is odd. With this reasoning in mind, it will be convenient to consider the following cases separately.
Case 1. $M=2^{n}, n \in \mathbb{N} \backslash\{0\}$.
In this instance, consider $8 M=2^{n+3}=a b$, where $(a, b)=\left(2^{i}, 2^{n+3-i}\right)$ for $i=0,1, \ldots, n+3$ with $a+b=2\left(2^{i-1}+2^{n+2-i}\right)=2(2 s+1)$ only when $i=1, n+2$. However, since both factorizations are equivalent, we need only investigate the solution of \eqref{1} when $(a, b)=\left(2,2^{n+2}\right)$. Thus, one finds
that $(X, Y)=\left(1+2^{n+1}, 2^{n+1}-1\right)$ and so $(x, y)=\left(2^{n}, 2^{n}-1\right)$. Hence, there exists only the trivial representation $M=T(M)-T(M-1)$.
Case 2. $M \neq 2^{n}$.
Clearly, $M=2^{m}(2 n+1)$ for an $n \in \mathbb{N} \backslash\{0\}$ and $m \in \mathbb{N}$. However, as there are more available factorizations of $8 M$, due to the presence of the term $2 n+1$, it will be necessary to consider the following subcases based on the possible factorizations $c d=2 n+1$.
Subcase 1. $(c, d)=(1,2 n+1)$.
Here consider $8 M=2^{m+3}(2 n+1)=a b$, where $(a, b)=\left(2^{i}, 2^{m+3-i}(2 n+1)\right)$ for $i=0,1, \ldots, m+3$ with $a+b=2\left(2^{i-1}+2^{m+2-i}(2 n+1)\right)=2(2 s+1)$ only when $i=1, m+2$. Solving (1) with $(a, b)=$ $\left(2,2^{m+2}(2 n+1)\right)$, one finds that $(x, y)=\left(2^{m}(2 n+1), 2^{m}(2 n+1)-1\right)$, which corresponds to a consecutive triangular number difference of $M$, while for $(a, b)=\left(2^{m+2}, 2(2 n+1)\right)$ we have $(x, y)=$ $\left(2^{m}+n, n-2^{m}\right)$ and so
$$
M=T\left(2^{m}+n\right)-T\left(y^{\prime}\right),
$$
where $y^{\prime}=2^{m}-n-1$ if $y<0$ and $y^{\prime}=y$ otherwise. In either situation, one has $\left|x-y^{\prime}\right|>1$, giving a nonconsecutive triangular number representation of $M$.
Subcase 2. $(c, d), c \neq 1,2 n+1$.
Here consider $8 M=2^{m+3} c d=a b$, where $(a, b)=\left(2^{i} c, 2^{m+3-i} d\right)$ for $i=0,1, \ldots, m+2$ with $a+b=2\left(2^{i-1} c+2^{m+2-i} d\right)=2(2 s+1)$ when $i=1, m+2$. Solving (1) with $(a, b)=\left(2 c, 2^{m+2} d\right)$, one has $(X, Y)=\left(c+2^{m+1} d, 2^{m+1} d-c\right)$, from which it is immediate that
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realnumber
Posted 2019-9-8 23:28
青青子衿
Posted 2019-9-9 09:17
hbghlyj
Posted 2022-8-5 05:49
