Proving an integer is composite

  • Thread starter Thread starter ver_mathstats
  • Start date Start date
  • Tags Tags
    Composite Integer
Click For Summary
SUMMARY

The discussion focuses on proving that at least one of the integers in the set defined by the expressions 9a+1, 9a+10, 9a+19, and 9a+28 is composite. Participants suggest using proof by contradiction and exploring the evenness or oddness of the expressions based on the parity of 'a'. Key insights include the importance of examining even numbers and their divisibility by 4 to identify composite numbers. The conversation emphasizes generating values for 'a' to systematically explore the properties of the sequences.

PREREQUISITES
  • Understanding of composite and prime numbers
  • Familiarity with the Euclidean Algorithm
  • Basic knowledge of mathematical proofs, particularly proof by contradiction
  • Ability to analyze sequences and modular arithmetic
NEXT STEPS
  • Explore the properties of composite numbers and their factors
  • Learn about generating sequences and their implications in number theory
  • Study proof techniques, especially proof by contradiction
  • Investigate modular arithmetic and its applications in number theory
USEFUL FOR

Mathematicians, students studying number theory, and anyone interested in understanding proofs related to composite and prime integers.

ver_mathstats
Messages
258
Reaction score
21
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.
 
Physics news on Phys.org
Are there even numbers among them?
 
  • Like
Likes   Reactions: scottdave, topsquark and PeroK
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?
 
  • Like
Likes   Reactions: ver_mathstats and topsquark
WWGD said:
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
 
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.
 
  • Like
Likes   Reactions: ver_mathstats and topsquark
ver_mathstats said:
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.
 
  • Like
Likes   Reactions: topsquark
PeroK said:
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
 
ver_mathstats said:
i am not the best with proofs unfortunately
can you prove that one of ##9a+1## and ##9a +2## is composite?
 
  • Like
Likes   Reactions: topsquark
Ok, I guess I overcomplicated it. @ver_mathstats : Have you tried to just generate quartets using different values of a?
 
  • #10
WWGD said:
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
 
  • #11
ver_mathstats said:
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.
 
  • #12
ver_mathstats said:
yes I have, and i get some prime numbers
Have you found ##a## where all four numbers are prime?
 
  • Like
Likes   Reactions: WWGD and scottdave
  • #13
For a given value of a, look at two cases: one where a is even and the other where a is odd.
 
  • Like
Likes   Reactions: scottdave and malawi_glenn
  • #14
ver_mathstats said:
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?
 
  • Like
Likes   Reactions: WWGD, scottdave and PeroK

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
18
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K