Using results by Andrew Wiles is allowed
x z + y z = 5 z
If x , y , z are positive integers, find the total number of ordered triplets ( x , y , z ) that satisfy the equation above.
Inspired by this problem .
The answer is 6.
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2 solutions
Moderator note:
Nicely done, unfortunately we need to implore Fermat's last theorem to solve this problem. Bonus question: What would the answer be if z is still an integer but not necessarily positive?
To apply FLT, x , y , 5 could also be any nonzero integers. FLT can easily be extended: x n + y n = z n has no solutions when n ≥ 3 , x y z = 0 . I don't like how wikipedia says 'positive a , b , c ' instead of 'nonzero integers a , b , c '.
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It can be extended, but the extension is rather trivial (if n is odd, just move those with negative terms to the other side; if n is even, it doesn't matter that they are negative since raised to an even power they become positive). That's why stating for positive integers alone is enough.
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well yes, it is trivial, but you'll have to prove it every time you want to use FLT for nonzero integers. You'll have to provide a full proof of the extension and only citing FLT may not be enough because of the precise statement being different.
Challenge Master:
Without FLT: Note that x , y ≤ 4 , so x z + y z ≤ 2 ⋅ 4 z . Thus 5 z ≤ 2 ⋅ 4 z , or z ≤ lo g 1 . 2 5 2 . For integer z , this gives z ≤ 3 , for a total of 1 2 cases to check.
For real z , this is much harder, since z < 0 is possible. With z < 0 , we obtain x , y > 5 instead of x , y < 5 , and this case has an unbounded number of cases (in fact, I think for each possible x , y there is at least one z ).
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Oh, I meant to say that z is not necessarily positive but still an integer. I've updated my note. Can you update your answer too?
I wasn't expecting a non-FLT solution. You should post that in your official solution!
As,solved below, for z>0, answer is 6. For z=0, LHS = 2, RHS = 1, so z=0 is ruled out. Now let z = -p. Then, 1/(x^p) + 1/(y^p) = 1/(5^p) or, (x^p) + (y^p) = (x^p) (y^p)/(5^p) or, (x^p) + (y^p) = (x y/5)^p LHS is integer, so RHS has to be an integer. Only possible if xy/5 is integer. Again using FLT, no solution for p>2. Now, for p = 1, x + y = x y/5 or, (x-5) (y-5)=25 Solutions : (6,30),(30,6),(10,10). For p =2, x^2 + y^2 = (x*y)^2/25 or, (x-25)(y-25)=625 No solution. So , 9 solutions in total.
If z ≥ 3 , by Fermat's Last Theorem , there is no solution, since x , y , 5 are positive integers. Thus z = 1 , 2 . Also, x z = 5 z − y z < 5 z , so x < 5 , thus x = 1 , 2 , 3 , 4 . By checking all eight cases, we have 6 solutions, ( x , y , z ) = ( 1 , 4 , 1 ) , ( 2 , 3 , 1 ) , ( 3 , 2 , 1 ) , ( 4 , 1 , 1 ) , ( 3 , 4 , 2 ) , ( 4 , 3 , 2 ) .
While at that, here's a non-FLT solution:
Note that x z , y z > 0 for being a positive number raised to the power of a positive number. Thus x z = 5 z − y z < 5 z , and likewise y z < 5 z . Because z > 0 , t h e f u n c t i o n t ↦ t z is strictly increasing on the positive reals, so x z < 5 z implies x < 5 , and likewise y < 5 .
Now, since x , y are integers, x , y < 5 implies x , y ≤ 4 . Thus 5 z = x z + y z ≤ 4 z + 4 z = 2 ⋅ 4 z , or ( 4 5 ) z ≤ 2 , or z ≤ lo g 5 / 4 2 . Apparently, lo g 5 / 4 2 < 4 , so z < 4 or z ≤ 3 . Thus we have 12 cases ( x = 1 , 2 , 3 , 4 and z = 1 , 2 , 3 ).