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 2020-03-20, 14:35 #1 enzocreti   Mar 2018 10138 Posts 216 is the only cube that can be written in this way? 216=3^2*(182^2-331*10^2) is there any other positive cube that can be written as a^2*(b^2-r*10^2)? with a and b positive and coprime, r prime?
2020-03-20, 15:16   #2
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

24·641 Posts

Quote:
 Originally Posted by enzocreti 216=3^2*(182^2-331*10^2) is there any other positive cube that can be written as a^2*(b^2-r*10^2)? with a and b positive and coprime, r prime?
Since d^3 = d^2 * d, all you are asking is whether d = (b^2 - 100r) is positive. In other words, for what values of b and r is b^2 > 100r and b is co-prime to d..

It should now be obvious that there are an infinite number of solutions. Picking just one pretty much at random, let r =1. In that case b^2 > 100, or b> 10. One such value is 11. Plugging in the numbers, d=121-100 = 21.

Indeed 21^3 = 9261 = 441 * (11^2 - 10 * 1^2).

Why do you post questions which are so trivial to answer with even the slightest amount of thought?

Last fiddled with by xilman on 2020-03-20 at 15:20

2020-03-22, 11:13   #3
LaurV
Romulan Interpreter

Jun 2011
Thailand

2·3·31·47 Posts

Quote:
 Originally Posted by xilman r =1
Neee.. gotcha! (so long I waited for that! )
r must be prime, I guess you didn't know that 1 is not considered a prime

2020-03-22, 15:30   #4
xilman
Bamboozled!

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May 2003
Down not across

24×641 Posts

Quote:
 Originally Posted by LaurV Neee.. gotcha! (so long I waited for that! ) r must be prime, I guess you didn't know that 1 is not considered a prime
Fairy Nuff: I over-looked that condition.

OK, choose r=2. Prime, right?

Let b=17, so B^2-100r = 289-200 = 89 = a. All are positive and co-prime.

Let b=19, so b^2-100r = 361-200 = 161 = 7 * 23 = a. All are positive and co-prime.

Let b=21, so a=241 which is a prime.

And so on.

It is still a trivial problem.

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