Was sorting through my image folder to do some cleanup and came across this. I don't know how to solve it.I broke it into cases and was doing pretty well until I got into ''n''=4''i''+1 for natural numbers ''i''. I eventually worked out that I would

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File: 1426561666932.jpg (70.58 KB, 937x960, 937:960, 1394243539212.jpg)

03/17/15 (Tue) 03:07:47 No.1979

Was sorting through my image folder to do some cleanup and came across this. I don't know how to solve it.

I broke it into cases and was doing pretty well until I got into n=4i+1 for natural numbers i. I eventually worked out that I would have a solution if I could prove that for any natural i, there would be natural numbers a and j such that 4j-1 | 12i + 13 + 4a, but that looks stupid as hell and makes me feel like I've gone wrong somewhere.

If there's an easier way to do this, I'd appreciate being shown this solution. I know I've been trying to find a formula for the solutions rather than just verifying they can be found, so I may be making the problem harder than it is.

As far as the other cases go, a=1 and b = 4 for n = 4i, b = 2 and a = i for n = 4i - 2, and b = 1 and a = i for n = 4i - 1.

03/17/15 (Tue) 08:46:46 No.1981

n=5 is a counterexample

03/18/15 (Wed) 19:27:10 No.1983

Excuse my ignorance, but what does that bar separating a, n and b, n mean?

03/19/15 (Thu) 19:11:48 No.1986

>>1983
a|b (a divides b) iff there is an integer k such that b=ka. In other words, b/a has no decimal part.

03/19/15 (Thu) 22:50:45 No.1989

>>1981

It's not. a = 1 and b = 3.

In fact, if n is of the form 12i-7 for natural numbers i, a = 1 and b = 3 is always a solution. If n is of the form 12i-3, then a = 3 and b = 3 is always a solution. When n is of the form 12i + 1, I'm stumped.

03/20/15 (Fri) 01:01:27 No.1991

>>1981
>>1989

To add, here are some valid solutions for the first few values of n that I can't prove all have solutions.

n = 13, (a, b) = (2,3)
n = 25, (a, b) = (5,15)
n = 37, (a, b) = (2,3), (5,3)
n = 49, (a, b) = (7, 7), (14,7)
n = 61, (a, b) = (2,3), (8,3)
n = 73, (a, b) = (1,7), (1,11)
n = 85, (a, b) = (2,3), (11,3), (5,15), (5,35), (17,51)
n = 97, (a, b) = (5,3), (2,7), (2,15)

05/29/15 (Fri) 07:49:26 No.2437

I ran a program and found out

n=60i-35 a=5 b=15

n=24i-11 a=2 b=3

n=84i-35 a=7 b=7

n=84i-11 a=1 b=7 (i is natural number)

these and the cases anon stated would satisfy all,I assume.

I think there should be the smarter way to solve this problem as OP said.

It is ridiculus that you must break it into 84 cases in order to solve the problem.

Sorry for bad English.