# another thing

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#1
prove that the difference of any two squares can be written as a sum of distinct consecutive odd numbers

(Im not sure this is true but I think it is)
0
15 years ago
#2
49 - 16 = 31
How can you write this as a sum of consecutive odd numbers?

Edit: for that matter, 9-4=5. 5 cannot be written as a sum of consecutive odd numbers.

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#3
(Original post by J.F.N)
49 - 16 = 31
How can you write this as a sum of consecutive odd numbers?

Edit: for that matter, 9-4=5. 5 cannot be written as a sum of consecutive odd numbers.

49 - 16 = 33*

also 5 is an odd number- in this sense you can consider a single odd number as a sum of one consecutive odd numbers. it is a problem with my wording of the problem.
0
15 years ago
#4
For any positive integers a > b,

(2b + 1) + (2b + 3) + (2b + 5) + ... + (2a - 1)
= (a - b) * (1/2)[(2b + 1) + (2a - 1)]
= (a - b) * (1/2)(2b + 2a)
= (a - b) * (a + b)
= a^2 - b^2
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15 years ago
#5
(Original post by mik1w)
49 - 16 = 33*
The prodigal mathematician. My apologies. I need to sleep.
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15 years ago
#6
I'm guessing that this comes from:
1² = 1
2² = 1+3
3² = 1+3+5
4² = 1+3+5+7
etc
I'll assume that you're talking about the absolute value of the difference. So say you had:
a² - b², a²>b²
You could write that as:
[1+3+5+...+(2A+1)] - [1+3+5+...+(2B+1)] for some A>B>0
= (2B+3) + (2B+5) + ... + (2A+1)
Which are consecutive odd numbers.
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#7
yup

prove 19*8^n + 17 is never prime for n>0.
I did not make this one up. can people do it in more than one method?
0
15 years ago
#8
19*8^(4k) + 17
1*(-1)^4k + 2 = 3 = 0 (mod 3)

19*8^(4k+1) + 17
19*8*4096^k + 17
6*(-5)*1^k + 4 = 0 (mod 13)

19*8^(4k+2) + 17
19*64*8^(4k) + 17
1*1*(-1)^4k + 2 = 3 = 0 (mod 3)

19*8^(4k+3) + 17
19*512*4096^k + 17
(-1)*2*1 + 2 = 0 (mod 5)
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#9
neat. how did you know which modulo to use?
(im not sure if the grammar in that sentence is right.. i mean how did you find x when you look at 19*8^(4k+n) + 17 = 0 (mod x)?)
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15 years ago
#10
Just a guess... No actually I worked out the prime factorizations of a few of them on the computer and noticed that when n=4k it was divisible by 3, n=4k+1 it was divisible by 13 etc. I guessed that this pattern would always hold and it turns out it does.
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#11
ah okay, well done
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