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How may one prove (or disprove) that an expression such as (8+ 9*m) where m=1,2,3,4 ...... can never be power of any integer?
Ajit
Ajit
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Thanks Gene.
The problem however remains as to how much one may try to obtain counter examples. For instance take the following problem: There are two positive single digit numbers (a, b with a<b)) neither being a factor of the other. Now there is only one pair such that (a + b*m) where m = 1,2,3,4.... is never a power of any whole number. Find a, b.
How many pairs may one try to get a counter example? Very tedious to get to the right pair simply by finding counter examples.
Ajit
The problem however remains as to how much one may try to obtain counter examples. For instance take the following problem: There are two positive single digit numbers (a, b with a<b)) neither being a factor of the other. Now there is only one pair such that (a + b*m) where m = 1,2,3,4.... is never a power of any whole number. Find a, b.
How many pairs may one try to get a counter example? Very tedious to get to the right pair simply by finding counter examples.
Ajit
Well, the counter example is the last choice. If you have tried everything you can to prove a proposition you have to think "maybe it isn't true." There is no limit to the number and not finding one doesn't prove that the proposition is true. After a while you go back and try some other attack to prove it is true. Most un-true ones, where they are true for some values and not others you can't "prove" true or false by math, logic or anything else because it is sometimes. Remember Godel. There are true things which cannot be proven true.
I haven't looked at the new problem yet. I suspect it is true but you may have to try all 19 relatively prime pairs, throwing them out when you get a power and going on to the next but it is more likely there is a way to severly limit the possibilities by math.
I haven't looked at the new problem yet. I suspect it is true but you may have to try all 19 relatively prime pairs, throwing them out when you get a power and going on to the next but it is more likely there is a way to severly limit the possibilities by math.
Well, it will ALWAYS be power of an integer :idea: :ajit said:
8 + 9(1) = 17^1
8 + 9(2) = 26^1
...
8 + 9(n) = [8 + 9(n)] ^1
Even if power is >1, then it seems easy:
8 + 9(13) = 5^3
8 + 9(56) = 8^3
8 + 9(56) = 2^9 (Gene's)
With coprimes and 10>b>a>1, there's lots:
2 + 3(2) = 2^3
2 + 3(10) = 2^5
2 + 3(41) = 5^3
....
There's 75 cases keeping m < 100.
Gene says: "Well, the counter example is the last choice"
Denis says:"Well, the counter example is the FIRST choice!"
Why try and prove/disprove something when a counter example
(if there is some) can be quickly found by a simple computer program?
Hmmm, There are usually an awful lot of things to try for counter examples if the proposition is true. None of them will work while the correct proof will work the first time. Aristotal's program would still be running, trying for a/b=sqrt(2) while that it is not true requires only a few minutes and a few lines. I'm gullable and if asked to prove something my first assumption is that it is true.
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Hi Gene,
I'm back one more time.
Did exactly what you suggested and came down to two pairs, 3 & 7 and 6 & 8. In case of the former, one can write
3 + 7*312 = 2187 = 3^7
So we are left with just one set of numbers 6 & 8 and I have not been able to find a counter example yet. If anyone can, please write back so that we can disprove the propositon completely. If not then 6 & 8 is the required pair.
Ajit
I'm back one more time.
Did exactly what you suggested and came down to two pairs, 3 & 7 and 6 & 8. In case of the former, one can write
3 + 7*312 = 2187 = 3^7
So we are left with just one set of numbers 6 & 8 and I have not been able to find a counter example yet. If anyone can, please write back so that we can disprove the propositon completely. If not then 6 & 8 is the required pair.
Ajit
I miss read the problem, taking not being a factor as having no common factor.
My only real progress was that a cannot be a square because m=b+2sqrt(a) gives
a+b²+2b*sqrt(a)=(b+sqrt(a))² and you are way past that.
Knowing 6,8 is the answer IF the proposition is true may or may not be what the poser had in mind. I suspect that a proof that 6+8m is not a power is needed.
2(3+4m) is a power only if 3+4m is a multiple of 2 but (3+4m) is odd so it can't be. I think this is a valid proof. I believe it might show that 6,8 is indeed the only pair that it is true of. I have not found another pair that has the necessary properties, a is 2*odd and not a factor of b, b is 2 times even.
My only real progress was that a cannot be a square because m=b+2sqrt(a) gives
a+b²+2b*sqrt(a)=(b+sqrt(a))² and you are way past that.
Knowing 6,8 is the answer IF the proposition is true may or may not be what the poser had in mind. I suspect that a proof that 6+8m is not a power is needed.
2(3+4m) is a power only if 3+4m is a multiple of 2 but (3+4m) is odd so it can't be. I think this is a valid proof. I believe it might show that 6,8 is indeed the only pair that it is true of. I have not found another pair that has the necessary properties, a is 2*odd and not a factor of b, b is 2 times even.
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I think that does it. Thanks Gene
Ajit
Ajit