Vieta's take too long

$f(x)=10x^{10}+9x^9+....+2x^2+x=10(x-a_1)(x-a_2)...(x-a_{10})$ . remember this:the basic definition of a polynomial.we have the given expression equal to $100((i-a_1)(-i-a_1)(i-a_2)(-i-a_2)...(i-a_{10})(-i-a_{10}))=(10(i-a_1)(i-a_2)....(i-a_{10}))(10(-i-a_1)(-i-a_2)....(-i-a_{10}))$ this is equal to by our above def: $f(i)f(-i)$ it is not hard to compute, you can use AGP or differentiate GP. either ways, the answer will just be $(-6+5i)(-6-5i)=6^2+5^2=\boxed{61}$

(Adapted from Aareyan Manzoor's method. Rewritten for clarity.)

Notice that the given equation has a root $x=0$ , which can be ignored since in the given expression, $(0)^2+1 = 1$ , which is the multiplicative identity. Hence let $a_{10} = 0$ , and $\displaystyle f(x) = \sum_{m=1}^{10} mx^{m-1} = 10 \prod_{n=1}^9 (x-a_n)$ .

Applying the identity $(a+ib)(a-ib) = a^2+b^2$ , we can reform the given expression:

$\begin{aligned} 100 \prod_{n=1}^9 (a_n^2+1) &= 100 \prod_{n=1}^9 (a_n+i)(a_n-i) \\ &= \left[ 10 \prod_{n=1}^9 (-i-a_n) \right] \left[ 10 \prod_{n=1}^9 (i-a_n) \right] \\ &= f(-i)f(i) \end{aligned}$

As has been noted, the above may be computed easily. For variety's sake, I will present a method besides AGP/differentiating GP, which uses the aforementioned identity:

$\begin{aligned} f(-i)f(i) &= \left[ \sum_{m=1}^{10} m(-i)^{m-1} \right] \left[ \sum_{m=1}^{10} m(i)^{m-1} \right] \qquad (\text{Split into real and imaginary parts}) \\ &= \left[ \sum_{m=0}^4 (2m+1)(-i)^{2m} + \sum_{m=0}^4 (2m+2)(-i)^{2m+1} \right] \left[ \sum_{m=0}^4 (2m+1)i^{2m} + \sum_{m=0}^4 (2m+2)i^{2m+1} \right] \\ &= \left( \sum_{m=0}^4 (2m+1)(-1)^m \right)^2 + \left( \sum_{m=0}^{4} (2m+2)(-1)^m \right)^2 \\ &= 5^2 + 6^2 \\ &= \boxed{61} \end{aligned}$

Let $f(x)= \displaystyle \sum_{m=1}^{10} m x^m$ . The roots of this function are $a_1, a_2, \ldots,a_{10}$ and we can easily tell that none of the roots can be positive.

If we replace $x$ in the above function with $(-\sqrt{x-1})$ we get a new function $g(x)$ . For $g(x)=0$ the roots will be $a_1^{2}+1, a_2^{2}+1, \ldots,a_{10}^{2}+1$ .

Note:- since $\sqrt{a^{2}}=|a|$ and we know that roots of $f(x)$ are negative therefore the replacement we used is $(-\sqrt{x-1})$ .

Step 1 :- $g(x)=0$

Step 2:- $(-\sqrt{x-1})(1+3(-\sqrt{x-1})^{2}+5(-\sqrt{x-1})^{4}+7(-\sqrt{x-1})^{6}+9(-\sqrt{x-1})^{8})=-(2(-\sqrt{x-1})^{2}+4(-\sqrt{x-1})^{4}+6(-\sqrt{x-1})^{6}+8(-\sqrt{x-1})^{8}+10(-\sqrt{x-1})^{10})$

Step 3:- Squaring both sides and bringing all terms on one side we get a polynomial of degree 10.

Step 4:- Calculate the coefficient of $x^{10}$ . It is easy to calculate since only one term generates $x^{10}$ . It is $\boxed{100}$

Step 5:- Calculating the constant term in the polynomial. (Can be done easily because only few terms a constant.)

$(-1)(1-3+5-7+9)^{2}=(-2+4-6+8-10)^{2}+100x^{10})$ Note:- Here we are ignoring all the other terms as they are of no use.

$-25=36+100x^{10}$

$100x^{10}+61=0$ . Therefore the constant term is $\boxed{61}$

Step 6:- The product of all roots i.e $(a_1^{2}+1) (a_2^{2}+1) \ldots(a_{10}^{2}+1)$ of the polynomial is $\boxed{\frac{61}{100}}$

Step 7:- The required answer is 100 times the product of roots . Thus the answer is $\boxed{61}$ .

That was easy enough!!