Prime numbers and divisibility by 12

[Philip Robotic]

Homework Statement

Prove that if $p$ is a prime number and if $p>5$ then $p^2-37$ is divisible by $12$

Homework Equations

The Attempt at a Solution

So I think that the number $p^2-37$ should be expressed in a way that we can clearly see that it is divisible by 3 and by 2 twice (because $2\cdot 2\cdot 3=12$). I tried modifying the original expression into something like this: $$p^2-37=12k$$ Where $k$ is a natural number, but got stuck here.

I also tried doing something with this: $(p-\sqrt{37})\cdot (p+\sqrt{37})$ but what next?

I think I am missing a step where I could use some of the properties of prime numbers, but I have really no idea where and how. I've been trying to solve this task for a pretty long time already, unfortunately whiteout success.

 

[fresh_42]

If you write $p^2-37=12k$ a bit differently, i.e. move $36$ into the $12k$ term, then it's easier to apply the trick with the third binomial formula, and the result becomes obvious. Btw, it works for $p=5\,,$ too.

 

[WWGD]

I would suggest considering the two possible crimes in terms of $4 : 4k+1, 4k+3$ , or maybe easier conceptually but longer, the primes $12k+1, 12k+5, 12k+7, 12k-1$ (last is the same as $12k+11.)$

 

[PeroK]

Homework Statement

Prove that if $p$ is a prime number and if $p>5$ then $p^2-37$ is divisible by $12$

Homework Equations

The Attempt at a Solution

So I think that the number $p^2-37$ should be expressed in a way that we can clearly see that it is divisible by 3 and by 2 twice (because $2\cdot 2\cdot 3=12$). I tried modifying the original expression into something like this: $$p^2-37=12k$$ Where $k$ is a natural number, but got stuck here.

I also tried doing something with this: $(p-\sqrt{37})\cdot (p+\sqrt{37})$ but what next?

I think I am missing a step where I could use some of the properties of prime numbers, but I have really no idea where and how. I've been trying to solve this task for a pretty long time already, unfortunately whiteout success.

Just a couple of observations, to emphasise why the other comments above are so important.

You really should have noticed that $p^2 - 37$ being divisible by 12 is equivalent to $p^2 -1$ being divisible by 12. That's a big lesson to learn from this! Whenever you are looking at divisible by $n$, always think modulo $n$.

Second, how generally true is this? Are there other numbers for which $p^2 - 1$ is always divisible by them? You might like to try to find some. But, perhaps this is a very specific result. It doesn't work for $n = 10$ or $n = 14$ or $n =16$. Only $n = 12$. Although, actually, it works for $n = 24$ as well!

So, maybe a "proof" here is just working through the small number of options for primes module 12? This is called a proof by "exhaustion".

My point is that this is another important lesson: sometimes, especially in these sort of problems, the best solution may simply be to go through all the options.

 

[SammyS]

Homework Statement

Prove that if $p$ is a prime number and if $p>5$ then $p^2-37$ is divisible by $12$

It seems to me that two observations will essentially get you there.

1. Any prime number greater than 2 is an odd number. I.e. p = 2k + 1 .

2. Any prime number greater than 3 is not divisible by 3, so it's congruent to ±1 mod 3 .
​

 

[WWGD]

Just a couple of observations, to emphasise why the other comments above are so important.

You really should have noticed that $p^2 - 37$ being divisible by 12 is equivalent to $p^2 -1$ being divisible by 12. That's a big lesson to learn from this! Whenever you are looking at divisible by $n$, always think modulo $n$.

Second, how generally true is this? Are there other numbers for which $p^2 - 1$ is always divisible by them? You might like to try to find some. But, perhaps this is a very specific result. It doesn't work for $n = 10$ or $n = 14$ or $n =16$. Only $n = 12$. Although, actually, it works for $n = 24$ as well!

So, maybe a "proof" here is just working through the small number of options for primes module 12? This is called a proof by "exhaustion".

My point is that this is another important lesson: sometimes, especially in these sort of problems, the best solution may simply be to go through all the options.

True. I start getting upset when I hear , usually profs., rejecting solutions that are not " elegant ' enough. Hey, when solving hard problems becomes second nature I will start worrying about my solutions being elegant.