91 ^ (3) + 7 ^ (3) can be expressed as difference of at least one set of two perfect squares a and b i.e
91 ^ (3) + 7 ^ (3) = a ^ (2) - b ^ (2).
i need one value each for a and b also try for
91 ^ (11) + 7 ^ (11).
2006-07-01 19:51:09 · 9 answers · asked by rajesh bhowmick 2 in Science & Mathematics ➔ Mathematics
91 ^ (3) + 7 ^ (3) = 3895.5^ (2) - 3797.5 ^ (2)
i am only getting this value pls give the perfect squares.
2006-07-01 20:47:52 · update #1
ok jimbo in place of 7 if it is 6 then what will be the values of a and b
2006-07-01 21:47:34 · update #2
jimbo if u try a little bit more u will discover a great and beautiful relation of numbers that is odd powers with squares.
2006-07-01 21:50:13 · update #3
For the revised problem, with
"ok jimbo in place of 7 if it is 6 then what will be the values of a and b"
I get two solutions
a = 1126, b = 717 and
a = 3934, b = 3837
2006-07-02 04:29:05 · answer #1 · answered by Anonymous · 9⤊ 4⤋
Hi rajesh,
How are you doing?
I'll try to answer your question for "God's sake", LOL.
Any odd number, and hence any a^n+b^n where exactly one of a, b is odd, can be expressed as the difference of two squares in at least one way. But this is not necessarily true if both a and b are odd.
With your two sums you have a problem because 91^3 +7^3 is even and is divisible by 2, but not 4, since a+b = 7+91 = 98 and 98/2 = 49. The same is true of 91^11+7^11: it is divisible 2 but not 4
Now if you try to express either one of these sums, as a difference of two perfect squares say a^2 - b^2 then both a, b must be even since these sums are even. But that means both a^2 and b^2 are divisible by 4 and so your sums would have to be divisible by 4 and they aren't.
However, (91^3+7^3) /2 can be expressed as the difference of 2 squares.
(91^3+7^3)/2 = 376957 = 188479^2 - 188478^2
So 91^3+7^3 = 2(188479^2 - 188478^2)
I hope this helps.
PS Don't let the rude people bother you. They obviously couldn't answer your question.
[EDIT] Yes, for 6, we would have (as one solution):
91^3 + 6^3 = 376894^2 - 376893^2
91^11+ 6^11 = 1771843337437570314274^2 -
1771843337437570314273^2
My brain's a bit tired lol, so just tell me clearly about the relationship.
2006-07-01 21:24:14 · answer #2 · answered by Jimbo 5 · 0⤊ 0⤋
91^3 + 7^3 = (91 + 7)(91^2 - 91*7 + 7^2)
a^2 - b^2 = (a + b)(a - b)
Substituting doesn't help. It seems difficult.
2006-07-01 20:11:16 · answer #3 · answered by nayanmange 4 · 0⤊ 0⤋
shiish. Um somewhere out there in the universe this kind of mathematics might help you to get a life....
I know the answer, who doesn't duh,
I'll give you a hint......
Lifes to figgin short, close the book and quietly shove it up your math professors ****! squareded by unorthidox theta multiples of 33.
2006-07-01 19:59:57 · answer #4 · answered by The Magical Healer 1 · 0⤊ 0⤋
Check out these great sites. They can help you with your problem.
http://www.ams.org/online_bks/conm22/conm22-chIII.pdf
http://www.hermetic.ch/misc/numcol.htm
http://mathworld.wolfram.com/SquareNumber.html
http://home.att.net/~numericana/answer/numbers.htm
2006-07-01 20:18:35 · answer #5 · answered by Adyghe Ha'Yapheh-Phiyah 6 · 0⤊ 0⤋
sorry cant help u...im very poor at maths but yes i can pray to God that u would find ur answer...
2006-07-01 20:01:13 · answer #6 · answered by Simple gurl 4 · 0⤊ 0⤋
I think it is only trial-and-error procedure. Will you please tell me for what purpose you want to solve these? I am asking you out of curiosity only.
2006-07-01 20:05:21 · answer #7 · answered by K N Swamy 3 · 0⤊ 0⤋
go buy yourself a calculator for god's sake!
2006-07-01 19:54:38 · answer #8 · answered by Anonymous · 0⤊ 0⤋
dont frighten me....for god's sake!!!
2006-07-01 19:55:55 · answer #9 · answered by Anonymous · 0⤊ 0⤋