Verified answer

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