On an elementary two variable linear sum number theory integer value

Question

Suppose $a,b\in\Bbb N$ are odd coprime with $a,b>1$ then is it true that if $$2(x_1a+x_2b),\mbox{ }2(x_2a-x_1b)\in\Bbb Z$$ $$x_2\frac{(a+b)}2+x_1\frac{(a-b)}2,\mbox{ }x_1\frac{(a+b)}2-x_2\frac{(a-b)}2\in\Bbb Z$$ holds for some $x_1,x_2\in\Bbb R$ then $x_1,x_2\in\Bbb Z$ should hold?

Answer 1

Here's a proof, essentially along the lines suggested by didgogns . . .

Suppose

$$ax_1 + bx_2 = n_1\tag{eq1}$$

$$ax_2 - bx_1= n_2\tag{eq2}$$

$$ {\small{\left(\frac{a+b}{2}\right)}}x_1 + {\small{\left(\frac{a-b}{2}\right)}}x_2 = n_3 \tag{eq3} $$

$$ {\small{\left(\frac{a+b}{2}\right)}}x_2 - {\small{\left(\frac{a-b}{2}\right)}}x_1 = n_4 \tag{eq4} $$

$$\text{where}$$

\begin{align*} &\;\;\;\;\;{\small{\bullet}}\;\;a,b\;\text{are odd, coprime integers, with $a,b>1.$}\\[4pt] &\;\;\;\;\;{\small{\bullet}}\;\;x_1,x_2 \in \mathbb{R}.\\[4pt] &\;\;\;\;\;{\small{\bullet}}\;\;n_1,n_2,n_3,n_4 \in \mathbb{Z}. \end{align*}

Claim: $x_1,x_2 \in \mathbb{Z}.$

Let $I = \{n \in \mathbb{Z} \mid nx_1,nx_2 \in \mathbb{Z}\}$.

Our goal is to show $1 \in I$.

Note that $I$ is an ideal of $\mathbb{Z},$ so $I$ is principal.

Also, by Bezout'sTheorem, $u,v \in I \implies \gcd(u,v) \in I$.

Solving $ \text{(eq$1$)},\text{(eq$2$)} \;\text{for}\; x_1,x_2 $ yields

\begin{align*} x_1 &= \frac{an_1-bn_2}{a^2+b^2}\\[4pt] x_2 &= \frac{bn_1+an_2}{a^2+b^2} \end{align*}

so $a^2 + b^2 \in I$.

Let $d$ be the least positive integer such that $d \in I$.

Then $I = (d)$, so $I$ is the set of integer multiples of $d.$

In particular, $d\,{\mid}(a^2+b^2).$

Then

\begin{align*} &\gcd(a,b)=1\\[4pt] \implies\; &\gcd(a,a^2+b^2) = 1\;\,\text{and}\;\,\gcd(b,a^2+b^2) = 1\\[4pt] \implies\; &\gcd(a,d) = 1\;\,\text{and}\;\,\gcd(b,d) = 1 \end{align*}

Solving $ \text{(eq$1$)},\text{(eq$3$)} \;\text{for}\; x_1,x_2 $ yields

\begin{align*} x_1 &= \frac{-2bn_3+(a-b)n_1}{a^2-2ab-b^2}\\[4pt] x_2 &= \frac{2an_3-(a+b)n_1}{a^2-2ab-b^2} \end{align*}

so $a^2-2ab-b^2 \in I$. Then

\begin{align*} &a^2 + b^2 \in I\;\text{and}\;a^2-2ab-b^2 \in I\\[4pt] \implies\; &(a^2 + b^2) - (a^2-2ab-b^2) \in I\\[4pt] \implies\; &b(2a + 2b) \in I\\[4pt] \implies\; &d\,{\mid}b(2a + 2b)\\[4pt] \implies\; &d\,{\mid}(2a + 2b)\qquad\text{[since $\gcd(b,d)=1$]}\\[4pt] \implies\; &2a + 2b \in I \end{align*}

Sinilarly,

\begin{align*} &a^2 + b^2 \in I\;\text{and}\;a^2-2ab-b^2 \in I\\[4pt] \implies\; &(a^2 + b^2) + (a^2-2ab-b^2) \in I\\[4pt] \implies\; &a(2a - 2b) \in I\\[4pt] \implies\; &d\,{\mid}a(2a - 2b)\\[4pt] \implies\; &d\,{\mid}(2a - 2b)\qquad\text{[since $\gcd(a,d)=1$]}\\[4pt] \implies\; &2a - 2b \in I \end{align*}

Then

\begin{align*} &2a + 2b \in I\;\text{and}\;2a - 2b \in I\\[4pt] \implies\; &(2a + 2b) + (2a - 2b) \in I\\[4pt] \implies\; &4a \in I\\[4pt] \implies\; &d\,{\mid}4a\\[4pt] \implies\; &d\,{\mid}4\qquad\text{[since $\gcd(a,d)=1$]}\\[4pt] \implies\; &4 \in I \end{align*}

Since $a,b$ are odd, $a^2 + b^2 \equiv 1 + 1 \equiv 2 \pmod 4$, hence

\begin{align*} &a^2+b^2 \in I\;\text{and}\;4 \in I\\[4pt] \implies\; &\gcd(a^2+b^2,4) \in I\\[4pt] \implies\; &2 \in I\\[4pt] \end{align*}

Since $a,b$ are odd, $\large{\frac{a+b}{2}}$ and $\large{\frac{a-b}{2}}$ are integers, one of which is even, and the other odd.

If $\,\large{\frac{a+b}{2}}$ is even and $\large{\frac{a-b}{2}}$ is odd, then

\begin{align*} & \left(\small{\frac{a+b}{2}}\right)x_1 \in \mathbb{Z} \;\;\text{and}\;\; \left(\small{\frac{a+b}{2}}\right)x_2 \in \mathbb{Z} && \text{[since $\small{\frac{a+b}{2}}$ is even and $2 \in I$]} \\[4pt] \implies\; & \left(\small{\frac{a-b}{2}}\right)x_1 \in \mathbb{Z} \;\;\text{and}\;\; \left(\small{\frac{a-b}{2}}\right)x_2 \in \mathbb{Z} && \text{[by $(\text{eq}4)$ and $(\text{eq}3)$]} \\[4pt] \implies\; &\small{\frac{a-b}{2}} \in I\\[4pt] \implies\; &{\gcd}{\left(\small{\frac{a-b}{2}},2\right)} \in I\\[4pt] \implies\; &1 \in I && \text{[since $\small{\frac{a-b}{2}}$ is odd]} \end{align*}

Similarly, if $\,\large{\frac{a-b}{2}}$ is even and $\large{\frac{a+b}{2}}$ is odd, then

\begin{align*} & \left(\small{\frac{a-b}{2}}\right)x_1 \in \mathbb{Z} \;\;\,\text{and}\;\; \left(\small{\frac{a-b}{2}}\right)x_2 \in \mathbb{Z} && \text{[since $\small{\frac{a-b}{2}}$ is even and $2 \in I$]} \\[4pt] \implies\; & \left(\small{\frac{a+b}{2}}\right)x_1 \in \mathbb{Z} \;\;\,\text{and}\;\; \left(\small{\frac{a+b}{2}}\right)x_2 \in \mathbb{Z} && \text{[by $(\text{eq}3)$ and $(\text{eq}4)$]} \\[4pt] \implies\; &\small{\frac{a+b}{2}} \in I\\[4pt] \implies\; &{\gcd}{\left(\small{\frac{a+b}{2}},2\right)} \in I\\[4pt] \implies\; &1 \in I && \text{[since $\small{\frac{a+b}{2}}$ is odd]} \end{align*}

Thus, in both cases, $1 \in I$.

It follows that $x_1,x_2 \in \mathbb{Z}$, as was to be shown.

Answer 2

By OP, we have:$$x_1a+x_2b \in \mathbb Z\tag{1}$$$$x_2a-x_1b \in \mathbb Z\tag{2}$$$$x_1\frac{a+b}{2}+x_2\frac{a-b}{2} \in \mathbb Z\tag{3}$$$$x_2\frac{a+b}{2}-x_1\frac{a-b}{2} \in \mathbb Z\tag{4}$$ From $a(1)-b(2)$ and $b(1)+a(2)$, we have$$(a^2+b^2)x_i \in \mathbb Z\tag{5}$$From $2(3)-(2)+(1)$ and $2(4)+(1)+(2)$, we have$$2(a+b)x_i\in \mathbb Z\tag{6}$$

Since $a$ and $b$ are odd numbers, $a^2+b^2\equiv2\pmod4$ and $2(a+b)$ is multiple of $4$. It follows that$$(a+b)x_i\in \mathbb Z\tag{7}$$and from $(a+b)(7)-(5)$,$$2abx_i\in \mathbb Z\tag{8}$$

Now, it is easy to see that since $a$ and $b$ are coprimes, $ab$ and $a^2+b^2$ are also coprimes. Therefore, $\gcd(2ab,a^2+b^2)=2$. Now from $(5)$ and $(8)$, $$2x_i\in \mathbb Z\tag{9}$$

Observe that exactly one of $\frac{a\pm b}{2}$ is even and another is odd. Therefore, one of $x_1\frac{a+b}{2}$ and $x_2\frac{a-b}{2}$ is integer by $(9)$. By $(3)$, another is also integer. By $(9)$ and the fact that gcd of odd number and $2$ is $1$, we have at least one of $x_i$ is integer. Another one is also integer by $(4)$.