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October 8[edit]

Is there a set of functions that fulfil a certain condition[edit]

If I have a linear function in ONE variable, say 4j, for j an integer and j ≥ 1, then 4j = 4, 8, 12, 16 etc.

So if I want to get the values that 4j does NOT represent, I can use the set of functions 4j+n, where n = 1, 2, or 3.

But what if I have a linear function in TWO variables, say 4jk+j+k, for j and k integers, and j and k ≥ 1.

For j = 1 and k = 1, 2, 3, 4, etc., 4jk+j+k = 6, 11, 16, 21, etc.

For j = 2 and k = 1, 2, 3, 4, etc., 4jk+j+k = 11, 20, 29, 38, etc.

For j = 3 and k = 1, 2, 3, 4, etc., 4jk+j+k = 16, 29, 42, 55, etc.

Question: Is there a set of functions which give the values that 4jk+j+k does NOT represent? Bh12 (talk) 23:42, 8 October 2013 (UTC)[reply]

If j or k were allowed to be zero you get all the numbers, so any problem has to do with multiplying so you can think prime numbers. That isn't the answer directly but if you change your expression to ((4j+1)(4k+1)-1)/4 = n then you're asking which values of 4n+1 can't be expressed as (4j+1)(4k+1) when j and k are 1 or more. Dmcq (talk) 07:17, 9 October 2013 (UTC)[reply]

Yes, but is there a function or a set of functions that gives those values of n which cannot be expressed as ((4j+1)(4k+1)-1)/?Bh12 (talk) 14:37, 9 October 2013 (UTC)[reply]

I was hoping you'd figure that out for yourself. Anyway there are two sets of numbers of the form 4n+1 to think of. One starts 5,13,17,29,37,41,53. and the other 9,21,33,49,57,69,77 taking off 1 and diving by 4 gives 1,3,4,7,9,10,13 and 2,5,8,12,14,17,19 Dmcq (talk) 19:46, 9 October 2013 (UTC)[reply]

Figure it out for myself? You credit me with more intelligence than I have. Let's see - the first differences of the first function, or rather the first set, you give are 8,4,12,8,4,12. And the first differences of the second function/set you give are 12,12,16,8,12,8. From WHERE do you get these two functions/sets (with varying first differences for a supposedly linear expression)? — Preceding unsigned comment added by Bh12 (talk • contribs) 20:42, 9 October 2013 (UTC)[reply]

The first lot are the primes of the form 4n+1. The second lot are the product of two primes of the form 4n+3. I did say think prime numbers. Dmcq (talk) 22:21, 9 October 2013 (UTC)[reply]