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I find this beautiful...by Mathguy15
Tags: beautiful 
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#19
Feb1612, 04:28 PM

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P: 18,290

[tex]1999=1^3+321409^3+894590^3+(908211)^3[/tex] 


#20
Feb1612, 05:24 PM

P: 63




#21
Feb1612, 05:33 PM

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P: 18,290

I didn't even program it, I found this on a list online 


#22
Feb1612, 06:16 PM

P: 75

Hi Mathguy15,
Multiply out the terms, but first cube your 3rd term since you're going to cube it later anyway... n^3 + 30*n^2 + 300n + 1000 n^3 + 30*n^2  300n + 1000 (60*n^2)  1 → 30*n^2 + 30*n^2  60*n^2 = 0 1000 + 1000  1 = 1999 You could do the same thing for (n+8)^3 = 1n^3 + 24*n^2 + 192n + 512, and (n8)^3, but you'd have to replace (60*n^2) with (48*n^2) and 1999 with 1023. And so on... Note that you're not really dealing with cubes since they cancel out when you multiply out and collect the terms. 


#23
Feb1712, 09:37 AM

P: 81




#24
Feb1712, 10:26 AM

P: 75

If you want to generalize further mathguy15...
for any n or k in N First term: (1*k^0*n^3) + (3*k^1*n^2) + (3*k^2*n^1) + (1*k^3*n^0) = (n+k)^3 Second term: (1*k^0*n^3) + (3*k^1*n^2)  (3*k^2*n^1) + (1*k^3*n^0) Third term: (Derived from above 2nd terms doubled) Fourth term: 1 ... then figure out the formula for those coefficients (x) above (which is almost certainly related to the Binomial Coefficients aka "the coefficients of the expansion of (x+1)^n"). Do that and then you'll have a nice little equation in 3 variables that will work for any n, k or x in N. Binomial Coefficients: 1 1 1 > Powers of 1 1 2 1 > Powers of 2 1 3 3 1 > Powers of 3... 1 4 6 4 1 > Powers of 4... ...when inserted into Polynomials. In the example you posted, then set k = 10  AC 


#25
Feb1712, 02:58 PM

P: 75

Here's an example of what I was referring to in the previous post, thinkingwise. Take the form...
(10 + n)^5 + (10  n)^5 + (100*n^4) + (20000*n^2) + (1) = 199999 Echoing the form used by mathguy15... v = (10 + n) w = (10  n) x = (100*n^4)^(1/5) y = (20000*n'^2))^(1/5) z = 1 x = an integer for (100*(100*s^5)^4)^(1/5); n =(100*s^5) y = an integer for (20000*(125*s'^10)^2))^(1/5); n' = (125*s^10) Will have to see where to go from there, meaning what conclusions one might come to regarding: v^5 + w^5 + x^5 + y^5 + z^5 


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