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.