Minimum no of bits to represent X

Question

It took me much time to reach the solution where I find the value of X as 2 but still not sure whether this is correct or not. please help me with the solution

Answer 1

Let $Y = (61^{610}+1,61^{671}-1)^{61}$. Then, the expression becomes $$X=(Y\cdot Y^{10}+1,Y^{10}-1)$$Note that $Y\cdot Y^{10}+1=Y\cdot(Y^{10}-1)+Y-1$. And, $Y$ clearly has no common factors with either of the terms. So, this gcd is equivalent to $$(Y-1,Y^{10}-1)$$But, this is clearly equal to $Y-1$ as $Y-1|Y^{10}-1$. So, $X=Y-1$.

So we have $$X=(61^{610}+1,61^{671}-1)^{61}+1$$

EDIT

With Robert Israel's work, $Y=2^{61}$, so $X=2^{61}+1$, which clearly takes $62$ bits to express. And, we also note that since $2$ has a cyclicity of $4$, $2^{61}$ has a last digit of $2$, so $X\mod10=3$.

Answer 2

It turns out that $(61^{610}+1, 61^{671}-1) = 2$, while $(2^{671}+1, 2^{610}-1) =2305843009213693953 = 3 \cdot 768614336404564651$.

Note btw that the gcd of the polynomials $x^{671}+1$ and $x^{610}-1$ is $x^{61} + 1$, and $2305843009213693953 = 2^{61}+1$.