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Old 2016-03-19, 06:06   #1
PawnProver44
 
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Post Is there a prime of the form......

Is there a prime for every power of 3 just by adding 1 to a single digit. In other words, is there always a prime of the form 3^x+10^y for y ≤ x. (I found this to be false, More generally, is there a prime of the form a^x+b^y for fixed a, x, and b if a and b are coprime. For the example 3^x+10^y, I found this to be true for all x values less than 100:

3^1+10^1 is prime (13)
3^2+10^1 is prime (19)
3^3+10^1 is prime (37)
3^4+10^2 is prime (181)
3^5+10^4 is prime (10243)
3^6+10^1 is prime (739)
3^7+10^2 is prime (2287)
3^8+10^1 is prime (6571)
3^9+10^4 is prime (29683)
3^10+10^2 is prime (59149)
3^11+10^6 is prime (1177147)
3^12+10^8 is prime (100531441)
3^13+10^3 is prime (1595323)
3^14+10^6 is prime (5782969)
3^15+10^y is composite for y ≤ 15.

I tested x < 100

Any other known values of x such that this is true?
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Old 2016-03-19, 07:06   #2
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This holds for a random value of x too, take x=8084, hence 3^8084+10^2391 is prime! I don't see anything wrong with my thesis.
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Old 2016-03-19, 07:33   #3
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Quote:
Originally Posted by PawnProver44 View Post
This holds for a random value of x too, take x=8084, hence 3^8084+10^2391 is prime! I don't see anything wrong with my thesis.
"is prime!" will not do -- you need a proof
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Old 2016-03-19, 07:37   #4
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well the point is that for any x value (greater that 15) there should be a prime of the form 3^x+10^y, where y is less than x. A randomly picked value x=8084, should hold and in order to "fit in" the next value, 3^8084+10^2391 should be prime, which is just another pro for being prime. (I used a generator to find a proven prime)

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Old 2016-03-19, 07:53   #5
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Quote:
Originally Posted by PawnProver44 View Post
well the point is that for any x value (greater that 15) there should be a prime of the form 3^x+10^y, where y is less than x.
x=34, 62, 70, 80, 92, 100, 105, 113, 117, 120, 121, 123, 141, 145, 159, ...

What is the thesis again? Why do you believe it is true? What's the point?

For stuff like this where you're just stating random assertions with no basis, you probably don't need a proof (since finding one prime doesn't *mean* anything), but should at least say something like "is a BPSW PRP" or "passes BPSW, Frobenius-Underwood, and M-R with bases 3,5,7" or something. The latter is awfully convincing and only takes ~4 seconds for your x=8084 example. You can go do a real proof once there is a *point*.

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Old 2016-03-19, 09:38   #6
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A few x values picked at random:

48, 197, 387, 394, 466, 825,

Primes for each of them:

3^48+10^10
3^197+10^18
3^387+10^44
(none found)
3^466+10^186
(still none)

But still, a higher chance that there is a prime of the form 3^x+10^y (y less than x) for any x value.
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Old 2016-03-19, 10:18   #7
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What is the thesis again? Why do you believe it is true? What's the point?

For stuff like this where you're just stating random assertions with no basis
At this point, I'll take "expressing ones idea clearly". I think we're long ways away from "proving ones assertions" -- let's call it phase 2.
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Old 2016-03-19, 10:26   #8
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Proving some statements like mine seem hard. In general, giving examples supports your statements.
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Old 2016-03-19, 12:13   #9
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Quote:
Originally Posted by PawnProver44 View Post
Is there a prime for every power of 3 just by adding 1 to a single digit. In other words, is there always a prime of the form 3^x+10^y for y ≤ x. (I found this to be false, More generally, is there a prime of the form a^x+b^y for fixed a, x, and b if a and b are coprime. For the example 3^x+10^y, I found this to be true for all x values less than 100:
1) we know the two terms must be coprime so if a and b are coprime it fits our necessary condition however there are conditions on the exponents to stop the number from being composite so a and b being coprime are not sufficient.

in general mod 3 we have the following cases:

a=1 b=0-> sum of any powers mod 3 will be 1 mod 3

a=1 b=1-> sum of any powers mod 3 will be 2 mod 3

a=1 b=2-> sum will differ depending on y if y is odd the sum will be 0 mod 3, if y is even sum will be 2 mod 3.

a=2 b=0 sum will differ depending on x if x is odd the sum will be 2 mod 3, if x is even sum will be 1 mod 3.

a=2 b=1 -> sum will differ depending on x if x is odd the sum will be 0 mod 3, if x is even sum will be 2 mod 3.

a=2 b=2 -> sum now depends on both x and y:
if x and y are same parity ( both odd or both even) sum will be 2 mod 3.
if x and y are opposite parity the sum will be 0 mod 3.
a=0 b=0 -> divisible by 3 and not coprime so that's out

a=0 b=1 -> sum of any powers mod 3 will be 1 mod 3

a=0 b=2-> sum will differ depending on y if y is odd the sum will be 2 mod 3, if y is even sum will be 1 mod 3.

so this narrows down what x and y can be.

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Old 2016-03-19, 22:11   #10
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Proving some statements like mine seem hard. In general, giving examples supports your statements.
Thanks to the moderators for banning that person.

When entering a group (be it an Internet a forum or whatever...) one should observe first : how does the group communicate, interact ...

Before asking a question search to see if there is an obvious answer, of course when new, our search capabilities are not high and the response from the "establishment" should be tolerant.

Before posting and affirming something be sure you know what you talk about..

Before starting new subjects or theories...

And even knowing an trying to abide by those rules, there are post that shame us in retrospect.

Jacob

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