20100720, 14:49  #1 
May 2010
Prime hunting commission.
3220_{8} Posts 
Integers = sums of 2s and 3s.
I stumbled upon this one a while ago, as I posted elsewhere:
Initially, I thought it was only the primes that could be decomposed into 2s and 3s. Then, I realized I could generalize this to all integers greater than 3. Here are some examples: 47 = 2(19) + 3(3) 79 = 3(17) + 2(14) 106 = 3(20) + 2(23) 189 = 2(60) + 3(23) Etc. Proving it might be an easy task. Tips? Also: Congrats to me on a Fermat prime post. (257) > 2^(2^3)+1. Last fiddled with by 3.14159 on 20100720 at 14:49 
20100720, 14:54  #2  
Nov 2003
2^{2}·5·373 Posts 
Quote:
Let m and n be elements of Z+ such that (m,n) = 1. Ask yourself: What is the smallest integer M that is not representable as mx + ny??? This is a very well known problem. Google is your friend. 

20100720, 15:06  #3 
Aug 2006
2^{2}×1,493 Posts 
Searchbait: Sylvester, Frobenius, "happy meal"

20100720, 15:17  #4  
May 2010
Prime hunting commission.
2^{4}×3×5×7 Posts 
Quote:
The only counterexample I've found is 1. A probable counterexample is 4: 2+2, since there are no 3s in that decomposition. It's looking like the counterexamples are powers of 2: Let's test a few: 8 = 3(2) + 2 16 = 3(4) + 2(2). Powers of 3: 9: 2(3) + 3 27: 3(5) + 2(6). The only counterexamples I've found are {1, 4}. However, when using larger integer pairs, it fails horrendously: Let's use 6 and 7: 8 can't be represented as 6a + 7b 4 and 5: 10: 5+5 11: Cannot be expressed as 4a + 5b. The only pair for which it works so well is 2 and 3 (Also 1 and 2, the best pair of all.) 

20100720, 15:21  #5 
Aug 2006
2^{2}·1,493 Posts 

20100720, 15:22  #6  
(loop (#_fork))
Feb 2006
Cambridge, England
2·3,191 Posts 
Quote:
If it's greater than three and odd, then n3 is a multiple of two. This is not hard problem. 

20100720, 15:25  #7  
May 2010
Prime hunting commission.
2^{4}·3·5·7 Posts 
Quote:


20100720, 15:26  #8  
May 2010
Prime hunting commission.
2^{4}·3·5·7 Posts 
Quote:
109: 7(7) + 6(10) 110: 6(2) + 7(14) 111: 6(15) + 7(3) 112: 6(14) + 7(4) 113: 7(5) + 6(13) Etc. What's the final counterexample here? Last fiddled with by 3.14159 on 20100720 at 15:34 

20100720, 15:45  #9 
Aug 2006
2^{2}·1,493 Posts 
Last fiddled with by CRGreathouse on 20100720 at 15:45 
20100720, 16:18  #10 
Nov 2003
2^{2}·5·373 Posts 

20100720, 16:27  #11  
May 2010
Prime hunting commission.
2^{4}×3×5×7 Posts 
Quote:
Here it is. A better source. Last fiddled with by 3.14159 on 20100720 at 16:36 

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