if a, b, c and d are 4 different numbers, and a+1/b = b+1/c = c+1/d = d+1/a = x. What is x?

Question

Math question for us to ponder

Answer 1

a+1/b = x => a+1 = bx
b+1/c = x => b+1 = cx
c+1/d = x => c+1 = dx
d+1/a = x => d+1 = ax
=> a+b+c+d + 4 = x(a+b+c+d)
=>x = 1 + 4/(a+b+c+d)

Answer 2

a=1,b=2,c=3,d=4 and x=1

Answer 3

(a + 1)/b = (b + 1)/c = (c + 1)/d = (d + 1)/a = x

a + 1 = bx implies a = bx - 1 (Equ.1)
b + 1 = cx implies b = cx - 1 (Equ.2)
c + 1 = dx implies c = dx - 1 (Equ.3)
d + 1 = ax implies d = ax - 1 (Equ.4)

Plug b's value from Equ.2 into Equ.1 :
a = (cx - 1)x - 1 = cx^2 - x - 1 (Equ.5)

Plug c's value from Equ.3 into Equ.5 :
a = (dx - 1)x^2 - x - 1 = dx^3 - x^2 - x - 1 (Equ.6)

Plug d's value from Equ.4 into Equ.6 :
a = (ax - 1)x^3 - x^2 - x - 1 = ax^4 - x^3 - x^2 - x - 1

Rearrange : ax^4 - a = x^3 + x^2 + x + 1
Factorise : a(x - 1)(x + 1)(x^2 + 1) = (x + 1)(x^2 + 1)
Cancel similar terms : a(x - 1) = 1, or, x = 1 + 1/a

Working backwards with the initial equation,
we find that d = a, c = d, b = c and a = b,
which means that a, b, c and d must be the
same number, so the original premise is untrue.
The value of x is 1 + 1/n, where n is a, b, c or d,
and n can be any number.

Answer 4

One possible solution:
a = b = c = d = 1 and x = 2.

Answer 5

x = a + 1/b → bx = ab + 1
x = b + 1/c → cx = bc + 1
x = c + 1/d → dx = cd + 1
x = d + 1/a → ax = da + 1

(a + b + c + d)x = ab + bc + cd + da + 4

x = (ab + bc + cd + da + 4)/(a + b + c + d)

Answer 6

a+1=bx
b+1=cx
d+1=ax
c+1=dx

=> bx-a=cx-b=ax-d=dx-c=1

=> bx+b=cx+c=dx+d=ax+a=1

=> b(x+1)=c(x+1)=d(x+1)=a(x+1)=1

=> a=b=c=d=1/(x+1)

So in this way a,b,c,d shouldn't be diffrend numbers ! :-)