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# #Yup it's number theory

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(deleted)

Apr 17, 2020
edited by somehelpplease  May 8, 2020

#1
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a=1; b=1;c=gcd((a^3+b^3)/(a+b), (a*b)); if(c==9 and a-b==6, goto4, goto5);printc, a, b; a++;if(a<100, goto2, 0);a=1;b++;if(b<100, goto2, discard=0;

OUTPUT:  a = 9   and   b = 3

Apr 18, 2020
#3
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That was correct, thanks a bunch! Have a good day :D

#2
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I'm not too savvy with number theory, but maybe you could try using the euclidean algorithm with this problem?

Apr 18, 2020
#4
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I tried. . . believe me, I tried. Thanks for the suggestion anyway :D Have a nice day!

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Suppose that a and b are positive integers such that a-b=6 and  $$\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9$$ . Find the smallest possible value of b.

$$a-b=6\\ a=6+b$$

$$\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9\\ \text{gcd}\left(\frac{(a+b)(a^2-ab+b^2)}{a+b}, ab\right) = 9\\ \text{gcd}\left((a^2-ab+b^2), ab\right) = 9\\ \text{gcd}\left((a^2-2ab+b^2+ab), ab\right) = 9\\ \text{gcd}\left(((a-b)^2+ab), ab\right) = 9\\ \text{gcd}\left((6^2+ab), ab\right) = 9\\ \text{gcd}\left((36+ab), ab\right) = 9\\ \text{gcd}\left(9(4+\frac{ab}{9}), ab\right) = 9\\ \text{gcd}\left((4+\frac{ab}{9}), \frac{ab}{9}\right) = 1\\$$

So ab is a multiple of 9 AND ab is not even  AND   $$\frac{ab}{9}$$  is not even

If ab=9 then gcd is correct, but there are not values of a and b exactly 6 units apart

If ab=27 then gcd is correct, and 9*3=27,   9-3=6

So the smallest values of a and b are   9 and 3 respectfully.

The smallest value of b is 3.

LaTex:

\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9\\
\text{gcd}\left(\frac{(a+b)(a^2-ab+b^2)}{a+b}, ab\right) = 9\\
\text{gcd}\left((a^2-ab+b^2), ab\right) = 9\\
\text{gcd}\left((a^2-2ab+b^2+ab), ab\right) = 9\\
\text{gcd}\left(((a-b)^2+ab), ab\right) = 9\\
\text{gcd}\left((6^2+ab), ab\right) = 9\\
\text{gcd}\left((36+ab), ab\right) = 9\\
\text{gcd}\left(9(4+\frac{ab}{9}), ab\right) = 9\\
\text{gcd}\left((4+\frac{ab}{9}), \frac{ab}{9}\right) = 1\\

Apr 20, 2020
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Thanks a lot, Melody :D Have a good day!