# Help explain the notation $P_{\alpha - 3}(8y)^2$

Problem
Let $$P_n(x) = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x + \cdots + (-1)^n \dfrac{1.3.5 \cdots (2n - 3)}{2.4.6 \cdots 2n}x^n$$ Prove that if $\alpha = 8y + 1$ and $\alpha \geq 3$, then $P_{\alpha - 3}(8y)^2$ is a solution to $x^2 \equiv a \pmod{2^\alpha}$.

I'm really confused about that notation. Is $$P_{\alpha - 3}(8y)^2 = P_{\alpha - 3}(64y^2) \text{ or } P^2_{\alpha - 3}(8y) ?$$

Thanks

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You could always plug in some numbers and see which of the alternatives that seems to satisfy the statement that you're supposed to prove. (I'm too lazy to do that myself now.) :) – Hans Lundmark Apr 11 '11 at 8:24
This is certainly a good case for parentheses. As it is so easy to take the square inside in the first case, I'm with Gerry. – Ross Millikan Apr 11 '11 at 13:06
@Ross Millikan: Thank you. – Chan Apr 11 '11 at 17:36