Question

Find all pairs of points with integer coordinates in the x-y plane which lie on the line 3x+5y=13.

Answer 1

By inspection, can you find ANY one solution ?

Answer 2

One particular solution is \(x=1,\,y=2\).
\[
\begin{align*}
3(1+p)+5(2+q)&=13\\
3+3p+5+5q&=13\\
3p+5q&=0\\
p=-5k,&\,q=3k\\
x=1-5k,&\,y=2+3k
\end{align*}
\]
I originally tried to solve this and it doesn't make sense.
\[
3(1+p)+5(1+q)=13
\]

Answer 3

You're almost done right after finding one particular solution : (x, y) = (1, 2)

Answer 4

x = 1-5k
y = 2+3k

is the general solution! which part doesn't make sense ?

Answer 5

I substitute the wrong number and it does not make sense lol. Now it does!

Answer 6

This makes me feel like that I am fishing for medals lol.

Answer 7

lol no, diophantine equations are not so easy to make sense of when you initially start...

Answer 8

If you notice , the same story repeats in number theory, linear algebra and differential equations :

General solution = `particular solution` + `null solution`

which is kind of amazing..

Answer 9

This is because we are studying with linear Diophantine equation, linear differential equation and linear algebra?

Answer 10

Maybe.. i dont knw exactly but i think this `particular` + `null` stuff shows up in many other areas that involve solving equations..

Answer 11

here is another almost same way to look at your earlier solution :

\[3x+5y=13\]
By inspection \((x, y) = (1, 2)\) is a particular solution to the given eqn.
And the null solution is given by \(t(-5, 3)\) because \( 3(-5t) + 5(3t) = 0\)

So Genral solution = `particular solution` + `null solution` = \((1, 2) + t(-5, 3)\)

Answer 12

You don't need to do any work for finding the null solution as above trick works in general.

The null solution of equaiton \(ax+by = c\) is \(t(-b, ~a)\).

So all that matters in solving a linear diophantine equation is finding the particular solution.

Answer 13

That's one heck of a insight.