t=22*b+1; g=t/(b-1)?

What do You want exactly, Johnny? To solve this system in real numbers? In integer numbers? In natural numbers?

The first question is easy, You have almost nothing to do - take arbitrary b, not equal to 1, then find t = 22*b + 1, then find g = t/(b-1) and You are ready - the system has infinitely many solutions.

Much more interesting is to solve it in natural numbers. If g is to be integer, then t/(b-1) = (22b + 1)/(b-1) must be an integer but

(22b + 1)/(b-1) = (22b - 22 + 23)/(b-1) =

= (22(b - 1) + 23)/(b-1) = 22(b - 1)/(b-1) + 23/(b-1) =

= 22 + 23/(b-1), so 23/(b-1) must be an integer. But 23, being a PRIME NUMBER, can be factored only 23 = 1*23, so we have the following possibilities:

b - 1 = 1 or b - 1 = 23. That means b = 2 or b = 24, then

t = 22*2 + 1 = 45 or t = 22*24 + 1 = 529 and

g = 22 + 23/1 = 45 or g = 22 + 23/23 = 23.

Finally we found 2 natural solutions /there are no more!/:

b' = 2, t' = 45, g' = 45, and

b" = 24, t" = 529, g" = 23.

And if You wish to find all integer (event. negative or 0) solutions take into account that 23 = (-1)*(-23) so for b - 1 are two more possibilities: b - 1 = -1 or b - 1 = -23. This way You obtain the following:

b = 0, t = 1, g = -1 and b = -22, t = -483, g = 21 along with the previous two.

Hope that will satisfy You!

t= g(b-1)

+ t=22(b+1)

=>2t= g + 22

=> g = 2t - 22

t = .5g+11

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dont think u can get b, or integers for any of your unknowns for that fact