gcd (a,b) = 1 and a,b are greater than 1

[twisted_logic89]

If gcd (a,b) = 1 and a,b are greater than 1

Then if t is greater than a*b - a - b
t = ax + by (x,y are greater than or equal to zero)

and a*b - a - b +1 is the minimum

so... if you use 3 and 5 for the ab - a - b equation, you get 8.
then you can use a combination of the two numbers you used for a and b to create any number after the number you get from the ab-a-b equation.

the object is to prove this.

gcd (a,b) = 1
ar + bs = 1

then we will multiply that by 't'
a(tr) + b(ts) = t
one side would have to be negative for this to work

this must be disproved:
ab - a -b = a*n + b*m
where n,m are greater than or equal to 0

if x|yz, then x|y and x|z
gcd (x,y) = 1
so x|z

what would be the next step to take in solving this proof?

 

[daon]

Can you write out the full (exact) statement you're trying to prove?

 

[twisted_logic89]

If gcd (a,b) = 1 and a,b are greater than 1

Then if t is greater than a*b - a - b
t = ax + by (x,y are greater than or equal to zero)

and a*b - a - b +1 is the minimum


sorry, i know its a confusing mess....