Thread: possible values for simple equation

Express all possible positive integer values of m, n in terms of a third variable, k. (Use modular arithmetic)

Hello, DivideBy0!

We can solve this with "normal" algebra . . .

Express all possible positive integer values of in terms of a third variable,

Solve for .[1]

Since is an integer, must be a multiple of 4.
. . That is: . for some integer

Substitue into [1]: .

And we have parametric equations for all solutions:

. . . for any integer

Thanks, I have two question though
How would you express them if you could only have them as natural numbers, would you have 1 =< k =< 33?
and What are parametric equations? thanks again

Originally Posted by DivideBy0

Thanks, I have two question though
How would you express them if you could only have them as natural numbers, would you have 1 =< k =< 33?
and What are parametric equations? thanks again

As



, so for some

Then we have:

,

or:

.

Now if you choose any then:



is a solution to the original equation.

RonL

Hello, DivideBy0!

How would you express them if you could only have them as natural numbers?
Would you have: . ? . . . . Yes!

What are parametric equations?

A parameter is an "extra variable".

Instead of having as a function of ,

. . we can have: .

This opens the door to an entire universe of fascinating functions and graphs
. . . . . curves with loops, that intersect itself, etc.