Solving single equation with 2 variables

R

_rob

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Nov 28, 2018
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How do I go about solving this:
Find α, β ∈ ℤ such that 32α + 35β = 1
 
First solve the equation for one of the variables. In this case we can solve for \(\displaystyle \beta\), say. (Or solve for \(\displaystyle \alpha\). It doesn't matter in the end.)
\(\displaystyle 32 \alpha + 35 \beta = 1\)

\(\displaystyle \beta = \dfrac{1 - 32 \alpha}{35}\)

We can't do anything else, so we can say that \(\displaystyle \alpha\) can be anything and \(\displaystyle \beta = \dfrac{1 - 32 \alpha}{35}\) for any value of \(\displaystyle \alpha\). That's the best we can do.

-Dan
 
This is a linear Diophantine equation. Since gcd(32,35)|1, we know there are integral solutions, would begin by looking for a way to write 1 as a linear combination of 32 and 35.

[MATH]1=(20+3)(30+2)-(20+1)(30+5)=32(23)+35(-21)[/MATH]
Thus, the solutions can be given in the form:

[MATH](\alpha,\beta)=(35k+23,-32k-21)[/MATH] where \(k\in\mathbb{Z}\)
 
MarkFL said:
This is a linear Diophantine equation. Since gcd(32,35)|1, we know there are integral solutions, would begin by looking for a way to write 1 as a linear combination of 32 and 35.

[MATH]1=(20+3)(30+2)-(20+1)(30+5)=32(23)+35(-21)[/MATH]
Thus, the solutions can be given in the form:

[MATH](\alpha,\beta)=(35k+23,-32k-21)[/MATH] where \(k\in\mathbb{Z}\)
Click to expand...
How did you know it's a Diophantine equation?

-Dan
 
topsquark said:
I'm a Scientist! I look at all details! I'm trained to observe!

This is the second time I've done that today. ?

Thanks for the catch!

-Dan
Click to expand...

I have days like that too. Gotta love 'em. :)
 
topsquark said:
I'm a Scientist! I look at all details! I'm trained to observe!

This is the second time I've done that today. ?

Thanks for the catch!

-Dan
Click to expand...
I was trained as an historian. Most of the known details are irrelevant, and most of the relevant details are unknown. The combination of known irrelevancies and unknown relevancies provides delightfully numerous degrees of freedom.
 
Thanks for that. I found another solution that also seems to work
[MATH] 1=(1 \cdot 3)+(-1 \cdot 2) \\ = (-1 \cdot 32) + (11 \cdot 3) \\ = ( 11 \cdot 35 ) + ( -12 \cdot 32 ) [/MATH]
Which gives me:
α = -12+35n
β = 11-32n
n∈ ℤ

Is one solution preferred over the other?
 
Notice that 35k + 23 = 35k + 35 - 12 = 35(k + 1) - 12 = 35n - 12.

And - 32k - 21 = - 32k + 11 - 32 = 11 - 32(k + 1) = 11 - 32n.

These are simply equivalent descriptions of the same set.
 
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