Proving an integer is composite

[ver_mathstats]

Homework Statement:

Let a be an integer equal or greater than 0. Prove that at least one of the integers 9a+1, 9a+10, 9a+19 and 9a+28 is composite

Relevant Equations:

euclidean algorithm

I am really struggling in how to begin this problem. So far I have considered using the Euclidean Algorithm and trying to find the gcd of each number like gcd(9,10) but each time they give me 1 so that doesn't work. My next idea is to do a proof by contradiction where I start with assuming that the integers are prime and somehow find a contradiction but I am unsure how to proceed.

Any help would be appreciated as I am really unsure of how to proceed. Thank you.

 

[fresh_42]

Are there even numbers among them?

 

[WWGD]

There's some overlap. To the effects of the problem, all those are the same mod9:
9a+10=(9a+1)+9, etc.
Are you sure you're using the right numbers?

 

[ver_mathstats]

There's some overlap. To the effects of the problem, all those are the same mod9:
9a+10=(9a+1)+9, etc.
Are you sure you're using the right numbers?

Yes this is the question I was given

 

[Mark44]

Three of the numbers in the set 9a+1, 9a+10, 9a+19 and 9a+28 seem to me to be red herrings that aren't essential to the problem. The sequence $\{9a + 1 | a \ge 0\}$ contains as subsets the sequences that are generated by the other three expressions. A previous hint asking whether some members of the sequence are even might be helpful.

 

[PeroK]

Yes this is the question I was given

I'm struggling to see why you're struggling. It's difficult to give a hint without giving away the answer.

That said, the problem looks suspiciously easy.

 

[ver_mathstats]

I'm struggling to see why you're struggling. It's difficult to give a hint without giving away the answer.

That said, the problem looks suspiciously easy.

i am not the best with proofs unfortunately

 

[PeroK]

i am not the best with proofs unfortunately

can you prove that one of $9a+1$ and $9a +2$ is composite?

 

[WWGD]

Ok, I guess I overcomplicated it. @ver_mathstats : Have you tried to just generate quartets using different values of a?

 

[ver_mathstats]

Ok, I guess I overcomplicated it. @ver_mathstats : Have you tried to just generate quartets using different values of a?

yes I have, and i get some prime numbers

 

[fresh_42]

yes I have, and i get some prime numbers

It is not about getting prime numbers. You said that you were looking for the existence of composite numbers. Therefore, look at even numbers, and make sure they are at least divisible by 4. You only have to find some, neither all nor none.

 

[PeroK]

yes I have, and i get some prime numbers

Have you found $a$ where all four numbers are prime?

 

[Mark44]

For a given value of a, look at two cases: one where a is even and the other where a is odd.

 

[SammyS]

i am not the best with proofs unfortunately

Don't worry about a formal proof initially. Just play around with these expressions, somewhat systematically.

What can you say about the even-ness / odd-ness of $9a$ if $a$ is even?

What can you say about the even-ness / odd-ness of $9a$ if $a$ is odd?