# The unit digit of $\prod_{k=0}^{97}\left(2+\alpha_{k}^2\right)$, where $\alpha_0,\alpha_1,....,\alpha_{97}$ are the $98^{th}$ roots of unity

The unit digit of $$\prod_{k=0}^{97}\left(2+\alpha_{k}^2\right)$$, where $$\alpha_0,\alpha_1,....,\alpha_{97}$$ are the $$98^{th}$$ roots of unity

My Approach: Since $$\alpha_0,\alpha_1,....,\alpha_{97}$$ are $$98^{th}$$ roots of unity so $$z^{98}-1=0$$

Hence $$z^{98}-1=(z-\alpha_0)(z-\alpha_1)(z-\alpha_2).....(z-\alpha_{97})$$

Replacing $$z=2$$ didnot help me.

Then I wrote $$2+\alpha_{k}^2=(\sqrt2+i\alpha_k)(\sqrt2-i\alpha_k)=(-\alpha_k+i\sqrt2)(\alpha_k+i\sqrt2)$$

Hence $$\prod_{k=0}^{97}(2+\alpha_k^2)=\prod_{k=0}^{97}(i\sqrt2-\alpha_k)\prod_{k=0}^{n}(i\sqrt2+\alpha_k)$$

$$(i\sqrt2)^{98}-1=\prod_{k=0}^{97}(i\sqrt2-\alpha_k)$$

and $$(-i\sqrt2)^{98}-1=\prod_{k=0}^{97}(-1)^{98}(i\sqrt2+\alpha_k)$$

$$\implies \prod_{k=0}^{97}=\left((i\sqrt2)^{98}-1\right)\left((-i\sqrt2)^{98}-1\right)=(2^{49}+1)^2$$

Hence unit digit is $$9$$ but answer given is $$5$$

Where am I making mistake?

• Where is the answer given as 5? Commented May 3 at 9:01
• @BrianMoehring In my assignment Commented May 3 at 9:03
• Okay... "your assignment is wrong". Commented May 3 at 9:04
• Indeed, despite a tiny (local and harmless) mistake: $$(\sqrt2+i\alpha_k)(\sqrt2-i\alpha_k)=\color{red}-(-\alpha_k+i\sqrt2)(\alpha_k+i\sqrt2).$$ Commented May 3 at 9:23

Your answer is correct. Here's another way to see the same result:

Instead, note that $$x\mapsto x^2$$ takes the set of 98th roots of unity onto the set of 49th roots of unity and the map is two-to-one.

Therefore, your product is equal to $$\left(\prod_{k=0}^{48}(2+\beta_k)\right)^2$$ where $$\beta_0, \beta_1, \ldots, \beta_{48}$$ are the 49th roots of unity.

In particular, $$\left(\prod_{k=0}^{48}(2+\beta_k)\right)^2 = \left(\prod_{k=0}^{48}(-2-\beta_k)\right)^2 = \left((-2)^{49}-1\right)^2 = (2^{49}+1)^2.$$

Another way:

Let $$2+a_k^2=b_k$$ where $$0\le k\le97$$

If $$a_k,0\le k\le97$$ is a root of $$x^{98}-1=0,$$

$$b_k$$ will be roots of $$\displaystyle(y-2)^{49}=1\iff y^{49}-\binom{49}1y^{48}2^1+\cdots+\binom{49}{48}y(-2)^{48}+(-2)^{49}-1=0$$

By Vieta's formula, $$\prod_{k=0}^{48}b_k=-\dfrac{-(2^{49}+1)}1$$

As $$a_k,0\le k\le97$$ is a root of $$x^{98}-1=0$$ so will be $$-a_k$$ as $$(-x)^{98}-1=0$$

$$\prod_{k=0}^{97}(2+a_k^2)=\prod_{k=0}^{48}(2+a_k^2)(2+(-a_k)^2)=\left(\prod_{k=0}^{48}b_k\right)^2=?$$