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#1
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The prime decompositions of different integers m and n involve the same primes. The integers m+1 and n+1 also have this property. Is the number of such pairs (m,n) finite or infinite?
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#2
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Yes, the number of such pairs is finite or infinite...
but that's probably not what you meant [img]/images/graemlins/smile.gif[/img] |
#3
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Infinite.
If k is a positive integer, then pairs: [m,n] -> [ (2k)^2-1 , (2k+2)^2-1] [m+1, n+1] -> [(2k)^2 , (2k+2)^2] will have that required property of decomposition. |
#4
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This is not correct. For example, with k=5, you have m=99
and n=143 but 13 divides 143 but not 99. |
#5
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[ QUOTE ]
This is not correct. For example, with k=5, you have m=99 and n=143 but 13 divides 143 but not 99. [/ QUOTE ] They are both divided by 11. Just decompose it as a difference of squraes: a^2-b^2=(a-b)(a+b) |
#6
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[ QUOTE ]
[ QUOTE ] This is not correct. For example, with k=5, you have m=99 and n=143 but 13 divides 143 but not 99. [/ QUOTE ] They are both divided by 11. Just decompose it as a difference of squraes: a^2-b^2=(a-b)(a+b) [/ QUOTE ] 99 = 3 x 3 x 11 143 = 11 x 13 3 doesn't divide 143 and 13 doesn't divide 99 |
#7
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[ QUOTE ]
[ QUOTE ] This is not correct. For example, with k=5, you have m=99 and n=143 but 13 divides 143 but not 99. [/ QUOTE ] They are both divided by 11. Just decompose it as a difference of squraes: a^2-b^2=(a-b)(a+b) [/ QUOTE ] OP: [ QUOTE ] The prime decompositions of different integers m and n involve the same primes. The integers m+1 and n+1 also have this property. Is the number of such pairs (m,n) finite or infinite? [/ QUOTE ] Not just "one of the same prime", but (all of) the same primes. For example: m = 2, n = 8 m = 2^1 n = 2^3 ---> both share the same primes m+1 = 3, n+1 = 9 m+1 = 3^1 n+1 = 3^2 ---> both share the same primes |
#8
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Very likely infinite. If you believe the Lenstra-Pomerance-
Wagstaff conjecture about the number of Marsenne primes less than 2^x-1 being asymptotic to (e^gamma)(log_base2(x)), not only are there an infinity of Marsenne primes, but a function that describes asymptotically how many: http://en.wikipedia.org/wiki/Lenstra...aff_conjecture Now, suppose q is a Marsenne prime, i.e., q+1 = 2^p for some prime p>=2 and q is also a prime (an odd prime). With m=q-1, m+1=q, n+1=q^2, n=q^2-1=(q+1)(q-1)=(2^p)m. Then, m is even and clearly m and n have the same prime factors. Also, trivially, n+1 and m+1 have the same prime factors (namely, q). |
#9
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bigpooch,
I am looking for a solution using only high school math. |
#10
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Obviously, not many even looked at my post!
In my previous post, the requirement that q needed to be prime was unimportant. Let p>=2 be any integer. m = (2^p-2); m+1 = (2^p-1); thus, m is even n+1 = (m+1)^2 = (2^p-1)^2 n = (m+1)^2 - 1 = (m+2)m = (2^p)m so n has the same prime factors as m (since m is even). |
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