# Thread: Linear Diophantine Equation question

1. ## Linear Diophantine Equation question

hi everyone .

i need a little help in a direction of how to start this question...

"For what values of c does 2x + 5y = c have exactly one strictly positive integer solution?"

im not very good with applying constraints to things . ive been told that a little bit of enumeration will be needed in near the end to get the answer though.

thank you for your time!

2. Originally Posted by melissaanderson
hi everyone .

i need a little help in a direction of how to start this question...

"For what values of c does 2x + 5y = c have exactly one strictly positive integer solution?"

im not very good with applying constraints to things . ive been told that a little bit of enumeration will be needed in near the end to get the answer though.

thank you for your time!
Ok. Start with telling us why a LDE would only have one positive solution?

3. Thm: Given any $\displaystyle c\geq 0$, there exists exactly one solution $\displaystyle (x,y)$ such that $\displaystyle y=0$ or $\displaystyle y=1$. Proof: Given $\displaystyle c$, it is either even or odd. If $\displaystyle c$ is even then $\displaystyle (\frac{c}2,0)$ is a solution. If $\displaystyle c$ is odd, then $\displaystyle c-5$ is even, so $\displaystyle (\frac{c-5}2,1)$ is a solution. Call this initial solution $\displaystyle (x_0,y_0)$. (Note that if $\displaystyle c$ is odd but $\displaystyle c-5<0$, no positive solution exists, not even this initial one.)

Cor: Now for any $\displaystyle n=1,2,...$, $\displaystyle (x_0-5n,y_0+2n)$ is also a solution. To be a strictly positive solution, $\displaystyle x_0-5n\geq 0$, so $\displaystyle x_0\geq5$ for a second solution to exist.

Thus, for a second solution to NOT exist, $\displaystyle y_0=0$ or $\displaystyle y_0=1$, and $\displaystyle x_0<5$. So, consider the function $\displaystyle c(x,y)=2x+5y$ on the domain $\displaystyle (x,y)=(0 \to 4, 0 \to 1)$. These resulting values of $\displaystyle c$ are the ones for which only one positive $\displaystyle (x,y)$ solution exists.