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If 'N' be any number and a,b,c,... be its different prime factors then show that the sum of the squares of
continued...
all such numbers is
((N^3)/3)(1-1/a)(1-1/b)(1-1/c)...+(N/6)(1-a)(1-b)(1-c)...
Should we use euler's function to prove this problem.Plz anybody show me how to derive this.
dear friends here the sums of the squares of the numbers relatively prime(including unity) to and not exceeding 'N',must satisfy the above mentioned formula.
1 Answer
- 1 decade agoFavorite Answer
I don't know that much about number theory, but this formula appears not to work. Perhaps, you wrote down the wrong formula.
Lets examine several counterexamples.
6 = 2 * 3
2^2 + 3^2 = 4 + 9 = 13
((6^3)/3)(1 - 1/2)(1 - 1/3) = (216/3)(1/2)(2/3) = 72/3 = 24.
Since 13 does not equal 24, it clearly does not work for N = 6.
Now, suppose that N = 30.
N = 2 * 3 * 5
2^2 + 3^2 + 5^2 = 4 + 9 + 25 = 38
((30^3)/3)(1 - 1/2)(1 - 1/3)(1 - 1/5) = (9000)(1/2)(2/3)(4/5)
= 9000*(4/15) = 600*4 = 2400
Since 38 does not equal 2400, the formula also does not work for N = 30.
So, you may wish to double check your formula that you are trying to prove.