Are you smarter than me? 42

Roots of equation are ${ x }^{ 2 }-x-1$ are $\phi =\frac { 1+\sqrt { 5 } }{ 2 } ,\tau =\frac { 1-\sqrt { 5 } }{ 2 }$ .

As ${ x }^{ 2 }-x-1$ is factor of ${ zx }^{ 45 }+{ yx }^{ 44 }+1$ .

Therefore we get $z\phi ^{ 45 }+{ y\phi }^{ 44 }+1=0,{ z\tau }^{ 45 }+{ y\tau }^{ 44 }+1=0$

Subtracting 2nd equation from 1st we get $z\phi ^{ 45 }-{ z\tau }^{ 44 }+{ y\phi }^{ 44 }-{ y\tau }^{ 45 }=0=z(\phi ^{ 45 }-{ \tau }^{ 45 })+{ y(\phi }^{ 44}-{ \tau }^{ 44 })$

Now we have formula ${ f }_{ n }=\frac { \left( { \phi }^{ n }-{ \tau }^{ n } \right) }{ \sqrt { 5 } }$ where ${ f }_{ n }$ is $nth$ Fibonacci number.

Using above formula we have our expression as $z{ \sqrt { 5 } f }_{ 45 }+y{ \sqrt { 5 } f }_{ 44 }=0$ .,,,,,(1)

Dividing by $\sqrt { 5 }$ we get $z{ f }_{ 45 }+y{ f }_{ 44 }=0$ .

As we know $45th$ and $44th$ Fibonacci numbers are $1134903170$ and $701408733$ .

Now substituting in (1) we get our equation as $1134903170z+701408733y=0$

Clearly $z=\pm 701408733n,y=\mp 1134903170n$ for $n=0,1,2,....$

Our answer occur's at $n=1$ and $z=\pm 701408733$

So our answer is $\Huge\boxed{701408733}$ .

z x^45+y x^44+1=(x^2-x-1)(a43x^43+a42x^42+...+a1x+a0)

Hence,

a0=-1
-a0-a1=0 => a1=1
a0-a1-a2=0 => a2=a0-a1 = -1
a1-a2-a3=0 => a3=a1-a2 = 2
a2-a3-a4=0 => a4=a2-a3 = -3
....
a41-a42-a43=0 => a43=a41-a42=(-1)^44*fibonacci(43)

Now, y = a42-a43, z=a43

Hence abs(z) = abs(a43) = fibonacci(43) = 701408733