Find the triplet of positive integers that satisfy
x 3 = 3 y ⋅ 7 z + 8
Submit your answer as x + y + z .
Note
It's easy to try some values for x , y and z , and find out the integers that satisfy the equation... to show that is the unique triplet of positive integers, can be more complicated. There's a short, relatively, solution that makes use of Pell's equation; also, I've found one appyling basic modular arithmetic.
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Haha thanks. Certainly, I can say that it is original since I just inspired when solving problems about Pell's equation.
Your solution is great and is different to the solution that I had thought that applies Pell's equation. I will post it with the other solution.
x 3 − 2 3 = 3 y ∗ 7 z ⟹ 3 ∣ ( x − 2 ) ( x 2 + 2 x + 4 ) a n d 7 ∣ ( x − 2 ) ( x 2 + 2 x + 4 ) . I t i s o b v i o u s t h a t 3 ∣ ( x − 2 ) a n d 7 ∣ ( x 2 + 2 x + 4 ) . S i n c e 3 y ∗ 7 z + 8 i s o d d , x h a s t o b e o d d . S o x = 5 , 1 1 , . . . . I t c a n n o t b e 5 s i n c e m i n i m u m R H S = 3 1 ∗ 7 1 + 8 = 2 9 . T r y x = 1 1 . T h e n ( x 2 + 2 x + 4 ) = 1 2 1 + 2 2 + 4 = 1 4 7 = 3 ∗ 4 9 . a n d ( x − 2 ) = 9 . ∴ 3 y ∗ 7 z = 9 ∗ 3 ∗ 4 9 = 3 3 ∗ 7 2 . ⟹ ( x , y , z ) = ( 1 1 , 3 , 2 ) . ∴ x + y + z = 1 1 + 3 + 2 = 1 6 .
Uniqueness of solution has to be proved.
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First of all,what a beast!Congratulations on this problem,whether it's yours or not.My original solution solution is very long so i will skip through some parts of the solution hoping that someone could clarify it in the comments section.I'm really looking forward to the short solution you mentioned,although i have used pell's equation in the final step of my solution.Here we go!.It is easy to notice that x ≡ 2 ( m o d 3 ) . By some trivial use of LTE from the condition we stated above,we can say that v 3 ( x 3 − 2 3 ) = 1 + v 3 ( x − 2 ) ,we have that v 3 ( x − 2 ) = y − 1 ,that is x − 2 = 3 y − 1 ⋅ 7 α .(A bit of clarification needed here.)Substituting into the original equation it is easy to see that α = 0 .So x − 2 = 3 y − 1 Substituting back into the original equation and dividing by 3 y − 1 ,we get the equation: 3 2 y − 3 + 6 ⋅ 3 y − 2 + 4 = 7 z From this it follows that 1 2 ⋅ 7 z − 1 2 = m 2 ,for some positive integer m . Substitute m = 1 2 q ,to get the equation 7 z − 1 2 q 2 = 1 . .Reducing modulo 8,we arrive at the fact that z = 2 z 1 ,so we now have the pell equation k 2 − 1 2 q 2 = 1 ,where k = 7 z 1 .The fundemental solution is k = 7 . q = 2 ,from where we get x = 1 1 , y = 3 , z = 2 .Here comes the tricky part ,which for time reasons i will leave for others to prove.From the recurrence equation we have; x n + 1 = 7 x n + 2 4 y n y n + 1 = 1 2 x n + 7 y n .It can now be proved by induction that the highest power of 7 which divides x n is 1,and the uniqueness of the solution now follows.Moderators should feel free to edit my solution so that it more comprehensible to other people,because its very late in my country.