# Guess my birthday

This is just a funny question that I was elaborating... I know one way to solve (or maybe it's wrong...), but I want know if there is another way to solve this (when we keep adding conditions, there is an possibility of letting the exercise trivial and don't see this, I think...)

The month and day of my birthday are perfect squares and its product is an power of an prime, with positive exponent. Denote by $x$ the square root of the month and $y$ the square root of the day. Then

1. If you know the value of the product $xy$, you can certainly deduce the sum $x+y$.
2. Knowing the sum $x+y$, even knowing that condition 1 holds, doesn't exists the possibility of deduce the product $xy$.

The sum $x+y$ is relatively prime with the product of the month by the day.

(This isn't my real birthday, but I accept gifts =p)

And excuse my English...

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So, the month can only be either of January, April, or September, yes? The days are limited to $1$, $4$, $9$ and $16$; $25$ is inadmissible since that would go against the "prime power" assumption. – J. M. Aug 15 '12 at 2:23
BTW: consider the dates January 16 and April 4. – J. M. Aug 15 '12 at 2:41
And either the day or the month must be $1$, else $x+y$ would be divisible by the same prime that $xy$ is a power of. – Robert Israel Aug 15 '12 at 6:22
There are six possibilities (1/4, 1/9, 1/16, 1/25, 4/1, and 9/1) and all (up to permutation) give different sums, so I don't your point (2) makes any sense at all. Not to mention that nothing in here differentiates the day from the month, so it is rather ambiguous. – user641 Aug 15 '12 at 18:44
@SteveD: thanks for pointing that! =p but I forget to wrote: the point (2) holds considering only the point (1) and the first statement, the product be power of prime and day and month perfect square (in this case remains more possibilities), and I think (I hope =p) in this case (2) makes some sense – Yuki Aug 15 '12 at 18:58

The simpliest way to solve is to list all possibilities and just remove those not fitting.

• The month and day of my birthday are perfect squares

$$\small \begin{matrix}1.01 & 4.01 & 9.01 & 16.01 & 25.01 & 1.04 & 4.04 & 9.04 & 16.04 & 25.04 & 1.09 & 4.09 & 9.09 & 16.09 & 25.09\end{matrix}$$

• Product is a power of a prime

$$\small \begin{matrix} \textrm{day} & 4.01 & 9.01 & 16.01 & 25.01 & 1.04 & 4.04 &16.04 & 1.09 & 9.09 \\ \textrm{product} & 4 & 9 & 16 & 25 & 4 & 16 & 64 & 9 & 81 & \\ \textrm{prime power}& 2^2 & 3^2 & 2^4 & 5^2 & 2^2 & 2^4 & 2^6 & 3^2 & 3^4 \\ \end{matrix}$$

• Denote by x the square root of the month and y the square root of the day.

$$\small \begin{matrix} \textrm{day} & 4.01 & 9.01 & 16.01 & 25.01 & 1.04 & 4.04 &16.04 & 1.09 & 9.09 \\ \textrm{x} & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & \\ \textrm{y}& 2 & 3 & 4 & 5 & 1 & 2 & 4 & 1 & 3 \\ \end{matrix}$$

• If you know the value of the product $xy$, you can certainly deduce the sum $x+y$.

(which is: for given $xy$ there is only one value of $x+y$) $$\small \begin{matrix} \textrm{day} & 4.01 & 9.01 & 25.01 & 1.04 & 16.04 & 1.09 & 9.09 \\ xy & 2 & 3 & 5 & 2 & 8 & 3 & 9 & \\ x+y& 3 & 4 & 6 & 3 & 6 & 4 & 6 \\ \end{matrix}$$

• Knowing the sum $x+y$, even knowing that condition 1 holds, it isn't possibile to deduce the product $xy$.

(which is: for given $x+y$ there is more then one value of $xy$) $$\small \begin{matrix} \textrm{day} & 25.01 & 16.04 & 9.09 \\ x+y&6 & 6 & 6 \\ xy & 5 & 8 & 9 & \\ \end{matrix}$$

• The sum $x+y$ is relatively prime with the product of the month by the day.

$$\small \begin{matrix} \textrm{day} & 25.01\\ x+y&6 & \\ product & 5 \\ \textrm{GCD}& 1 \\ \end{matrix}$$

So the answer is January 25.

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