Yes but you can't say that 33 (mod 10) = 3*11 (mod 10) and therefore 33 mod 10 is divisible by 3 and 11 (but not 9 so it can't be a square, which is what the proof says).
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 81
 06012005 01:45

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 82
 06012005 01:52
(Original post by SsEe)
Yes but you can't say that 33 (mod 10) = 3*11 (mod 10) and therefore 33 mod 10 is divisible by 3 and 11 (but not 9 so it can't be a square, which is what the proof says).
43 = 33 (mod 10)... 
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 83
 06012005 13:02
Oh! Yeah, what you're saying makes perfect sense. The mixup was on my part. Here's another way to look at the question:
For n>4, S always ends in 3 and hence isn't a perfect square.
Proving that if a number ends in 3 it isn't a perfect square basically uses SsEe's method. 
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 84
 06012005 14:58
(Original post by dvs)
I don't get what you're saying. The proof isn't really dividing per se, it's considering the prime decomposition of 33 (mod 10).
for example 3=3.1 and 3=9.7 mod 10. 
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 85
 07012005 02:26
Even better and slicker, mod 5.

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 86
 07012005 19:09
Can 801,345,230,914 be written as the sum of 3 cubes? 
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 87
 07012005 22:29
Is that a guess or do you have some reasoning to go with that?

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 88
 07012005 22:36
I have some reasoning. (That I don't want to show in case I'm talking BS again. )

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 89
 07012005 22:39
Well let's see the reasoning then.

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 90
 07012005 22:39
Let's consider perfect cubes modulo 7:
a³ = b (mod 7)
0³=0, 1³=1, 2³=1, 3³=1, 4³=1, 5³=1, 6³=1 and 7³=0 (all mod 7)
So perfect cubes leave remainders 0, 1 or 1 when divided by 7. In other words, perfect cubes are of the form:
7k+1, 7k1 or 7k.
Now consider the sum of 3 cubes of the form 7k1, it'd be of the form 7p3=7n+4. So numbers of the form 7n+4 can be written as the sum of 3 perfect cubes. Let's check this for n={1,2,3}:
7*1+4 = 11 = 3³ + (2)³ + (2)³
7*2+4 = 18 = 3³ + (2)³ + (1)³
7*3+4 = 25 = 3³ + (1)³ + (1)³
Finally, let's consider 801,345,230,914 modulo 7:
801,345,230,914 = 4 (mod 7), which means it's of the form 7k+4 and can be written as the sum of 3 cubes. 
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 91
 08012005 00:01
Nice. Sadly you've only proved that a solution exists modulo 7.
I have another question.
a² + b² = 6ab. Find the value of (a+b)/(ab).
This is from an SMC so shouldn't be too hard. 
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 92
 08012005 00:11
a² + b² = (a+b)²  2ab = (ab)² + 2ab = 6ab
So (a+b)/(ab) = sqrt((a+b)²/(ab)²) = sqrt(8ab/4ab) = sqrt(2) 
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 93
 08012005 00:16
Well done . I knew the efforts of the TSR mathmos wouldn't fail to solve that really hard question hehe

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 94
 08012005 02:14
(Original post by SsEe)
Nice. Sadly you've only proved that a solution exists modulo 7.
How'd you do it? 
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 95
 08012005 02:49
You showed that numbers of the form 7k+4 can be written as 3 cubes modulo 7. In fact it can't be written as 3 cubes:
Consider modulo 9:
cubes modulo 9 can only be 0, 1 or 8.
If a solution exists:
a³ + b³ + c³ = 801,345,230,914
+(0, 1 or 8) + (0, 1 or 8) + (0, 1 or 8) = 4 (mod 9)
(+ because 1=8 and 8=1)
No combination of these make 4 mod 9 so no solution can exist.
If it was congruent to any of these there would be a solution mod 9 (but not necessarily mod 1) : 0, 1, 2, 3, 6, 7, 8 (I think that's all).
Somebody should check this as I'm having some doubts. 
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 96
 08012005 13:42
(Original post by SsEe)
You showed that numbers of the form 7k+4 can be written as 3 cubes modulo 7. In fact it can't be written as 3 cubes:
Consider modulo 9:
cubes modulo 9 can only be 0, 1 or 8.
If a solution exists:
a³ + b³ + c³ = 801,345,230,914
+(0, 1 or 8) + (0, 1 or 8) + (0, 1 or 8) = 4 (mod 9)
(+ because 1=8 and 8=1)
No combination of these make 4 mod 9 so no solution can exist.
If it was congruent to any of these there would be a solution mod 9 (but not necessarily mod 1) : 0, 1, 2, 3, 6, 7, 8 (I think that's all).
Somebody should check this as I'm having some doubts.
801 = 0 (mod 9), 345 = 3 (mod 9), so you cant say anything about these two... However,
230 and 914 = 5 (mod 9), but since cubes modulo 9 are 1,0,1, no sum of 3 can make 5, so these 2 cannot be written as the sum of 3 cubes. 
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 97
 08012005 13:45
(Original post by dvs)
But isn't that sufficient? I mean, I showed that numbers of the form 7k+4 can be written as the sum of 3 perfect cubes, and then showed that the number in the question is of the form 7k+4.
How'd you do it? 
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 98
 08012005 13:59
(Original post by JamesF)
No, it is not a sufficient condition. You have not proved that *every* single number which can be written in the form 7k+4 can be written as the sum of 3 cubes, only that the sum of 3 cubes will always be a number of the form 7k+4.
Rather he has shown that mod 7 considerations don't provide reasons why a number of the form 7k+4 couldn't be written as the sum of three cubes. 
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 99
 08012005 14:02
(Original post by RichE)
I don't think he's proven that, because it isn't true. e.g. 0^3 + 0^3 + 0^3.
Rather he has shown that mod 7 considerations don't provide reasons why a number of the form 7k+4 couldn't be written as the sum of three cubes. 
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 100
 08012005 14:13
345 = 7^3 + 1^3 + 1^3.
9^3 < 801 < 10^3, so a,b,c < 10
One of the cubes must be larger than 801/3 = 267, therefore
10 > a > 6
Then try a = 7,8,9
801  9^3 = 72,
801  8^3 = 289
801  7^3 = 458
Then its the same but showing each of those can or cannot be written as the sum of 2 cubes.
Starting with 72, 5 > b > 3, so b = 4, c = 2
So we have a solution
801 = 9^3 + 4^3 + 2^3
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Updated: August 3, 2015
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