Linear Diophantine Equation question

**melissaanderson** · January 18th 2010, 01:41 PM

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!

**Drexel28** · January 18th 2010, 05:30 PM

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?

**Media_Man** · January 18th 2010, 05:45 PM

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

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

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

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