# Thread: Birthday Problem; Prime Number Problem

1. ## Birthday Problem; Prime Number Problem

I am currently teaching myself some maths but I have found these two questions on the internet, and I'm completely stuck.

The questions are:

1.A man whose birthday was January 1st died on Feb 9th 1992. He was $x$ years of age in the year $x^2 AD$ for some interger $x$. In what year was he born?

2. If $p$ and $p^2+14$ are both prime numbers, find, with justification, all possible values of $p$.

I think that question 1 is to do with quadratics and you can tell that if $x$ is $44$, then $x^2$ would be $1936$ which looks to be the most sensible, although I do not know where to go. Question 2 I don't really know where to start.
These are both really nagging me and I cannot get the answers.

Thanks so much for any help given!

2. ## re: Birthday Problem; Prime Number Problem

Originally Posted by BobtheBob
I am currently teaching myself some maths but I have found these two questions on the internet, and I'm completely stuck.

The questions are:

1.A man whose birthday was January 1st died on Feb 9th 1992. He was $x$ years of age in the year $x^2 AD$ for some interger $x$. In what year was he born?

2. If $p$ and $p^2+14$ are both prime numbers, find, with justification, all possible values of $p$.

I think that question 1 is to do with quadratics and you can tell that if $x$ is $44$, then $x^2$ would be $1936$ which looks to be the most sensible, although I do not know where to go. Question 2 I don't really know where to start.
These are both really nagging me and I cannot get the answers.

Thanks so much for any help given!
2. Let $p$ be a prime other than 3.

$p\equiv\pm 1\pmod{3}$

$p^2\equiv 1\pmod{3}$

$p^2+14\equiv 1+14\equiv 0\pmod{3}$

$p^2+14$ is not prime.

When $p=3$, $p^2+14=23$ is a prime.

So, the only possible value of $p$ is 3.

3. ## re: Birthday Problem; Prime Number Problem

Originally Posted by BobtheBob
I am currently teaching myself some maths but I have found these two questions on the internet, and I'm completely stuck.

The questions are:

1.A man whose birthday was January 1st died on Feb 9th 1992. He was $x$ years of age in the year $x^2 AD$ for some interger $x$. In what year was he born?

2. If $p$ and $p^2+14$ are both prime numbers, find, witjustification, all possible values of $p$.

I think that question 1 is to do with quadratics and you can tell that if $x$ is $44$, then $x^2$ would be $1936$ which looks to be the most sensible, although I do not know where to go.
Good. You might also note that if x= 45, then $x^2= 2025$ which cannot be right because he died before that. If x= 43, then $x^2= 1849$ which would mean that when he died, in 1992, he was 43+ (1992- 1849)= 43+ 143= 186 years old! Now, if he was 44 in 1936, when was he born?

Question 2 I don't really know where to start.
p must not be 2 or 7 since, in that case, $p^2+ 14$ would be divisible by 2 or 7 respectively.

These are both really nagging me and I cannot get the answers.

Thanks so much for any help given!

4. ## re: Birthday Problem; Prime Number Problem

Question 1:

Observation 1: $\sqrt{1992} = 44.xxx$ hence maximum possible integers whose squares are less than 1992 are atmost 44. Now lets consider the simplest case of 1 for a man to be 1 year old on 1 AD he as to be born on 1 Jan 0 AD the he would be 1992 year of age when he died which would be absurd.

So for any $x^2$ AD the man has to be born $x^2 - x$ AD to be x years old and given that he died on 1992 only practical solution can be obtained if choose an year closer to 1992 so choosing the year 1936 we have $x^2 = 1936 \implies x = 44$. So the man should be born on 1936 - 44 = 1892 AD. So he was just over 100 years old when he died which is practically possible.

Kalyan.

5. ## re: Birthday Problem; Prime Number Problem

Ok thanks so much! I understand question 1. Is there a more elegant answer for question 1 or would trial and error be a good answer?

I'm still not all that sure on question 2 though. I don't quite understand why some steps are being done? for example:

$p\equiv\pm 1\pmod{3}$
surely $7 --1=8$ and therefore not divisable by 3?

Thanks very much!

6. ## re: Birthday Problem; Prime Number Problem

Originally Posted by BobtheBob
Ok thanks so much! I understand question 1. Is there a more elegant answer for question 1 or would trial and error be a good answer?

I'm still not all that sure on question 2 though. I don't quite understand why some steps are being done? for example:

surely $7 --1=8$ and therefore not divisable by 3?

Thanks very much!
$7\equiv 1\pmod{3}$

7. ## re: Birthday Problem; Prime Number Problem

Ok thanks very much for the help. I'm understanding it better now!

Thanks!

8. ## re: Birthday Problem; Prime Number Problem

Actually, are you able to explain this step for me because I don't quite understand what you are doing?

$p^2+14\equiv 1+14\equiv 0\pmod{3}$
Thanks very much.

9. ## re: Birthday Problem; Prime Number Problem

For any prime $p \not = 3,$
$p = 3k+1 \implies p^2 = (3k+1)^2 = 9k^2+6k+1 = 3K+1$
or
$p = 3k+2 \implies p^2 = (3k+2)^2 = 9k^2+12k+4 = 3K'+1$
So either ways
$p^2 = 3L+1$ for some $k,K,K',L \in Z^+$
so $p^2 \equiv 1 \mod(3) \implies p^2+ 14 \equiv 1+14 \equiv 0 \mod(3)$

Kalyan

10. ## re: Birthday Problem; Prime Number Problem

Ok thanks very much for all the help given!