Problem of the Day: June 12
Find all positive integers x, y, and z satisfying x2 + y2 + z2 = 2xyz.
(As usual, give solutions in comments to this post. On Friday I'll post some combination of my own solution and the most elegant approaches suggested by others.)
Edited to add solution: If x2 + y2 + z2 = 2xyz, then x2 + y2 + z2 must be even. Thus there are two cases to consider:
1. One of x, y, z is even and two are odd.
2. All three of x, y, and z are even.
Start with case 1. Assume (without loss of generality) that x is even, so x = 2a. Then we have 4a2 + y2 + z2 = 4ayz. But then y2 + z2 must be a multiple of 4. But odd numbers are always 1 mod 4, so this is impossible. Thus case 1 can't be achieved.
Now consider case 2. We have x=2a, y=2b, and z=2c. Thus 4a2 + 4b2 + 4c2 = 16abc, or a2 + b2 + c2 = 4abc.
So let's now focus on this new equation. Again we need the sum to be even, so we need either one or three even numbers. But one even and two odd won't work for the same reason as before. So we must have all three of a, b, and c even. Thus a=2d, b=2e, c=2f, and we have 4d2 + 4e2 + 4f2 = 32def, or d2 + e2 + f2 = 8def.
Now we're on to a new equation. In general, each step just pushes us to a new case. This requires us to get infinite sequences of even numbers -- but these numbers must be getting smaller with each step, which is impossible. Thus there are no solutions.
(Problem source: 2005 Morocco National Olympiad #2)
(As usual, give solutions in comments to this post. On Friday I'll post some combination of my own solution and the most elegant approaches suggested by others.)
Edited to add solution: If x2 + y2 + z2 = 2xyz, then x2 + y2 + z2 must be even. Thus there are two cases to consider:
1. One of x, y, z is even and two are odd.
2. All three of x, y, and z are even.
Start with case 1. Assume (without loss of generality) that x is even, so x = 2a. Then we have 4a2 + y2 + z2 = 4ayz. But then y2 + z2 must be a multiple of 4. But odd numbers are always 1 mod 4, so this is impossible. Thus case 1 can't be achieved.
Now consider case 2. We have x=2a, y=2b, and z=2c. Thus 4a2 + 4b2 + 4c2 = 16abc, or a2 + b2 + c2 = 4abc.
So let's now focus on this new equation. Again we need the sum to be even, so we need either one or three even numbers. But one even and two odd won't work for the same reason as before. So we must have all three of a, b, and c even. Thus a=2d, b=2e, c=2f, and we have 4d2 + 4e2 + 4f2 = 32def, or d2 + e2 + f2 = 8def.
Now we're on to a new equation. In general, each step just pushes us to a new case. This requires us to get infinite sequences of even numbers -- but these numbers must be getting smaller with each step, which is impossible. Thus there are no solutions.
(Problem source: 2005 Morocco National Olympiad #2)
Since x^2 + y^2 + z^2 is even, at least one of x, y and z is. Let x = 2a; then 2xyz = 4ayz, so 4a^2 + y^2 + z^2 = y^2 + z^2 = 0 (mod 4). Squares are either 0 or 1 mod 4, so y^2 and z^2 are both 0 mod 4, so y and z are even. Now we have 4a^2 + 4b^2 + 4c^2 = 16abc, a^2 + b^2 + c^2 = 4abc; and by the same argument, we can show that a, b, and c are even. In this way we can show x, y, and z are divisible by 2^n for any n; which means they are 0, which is not a positive integer. Therefore there are no solutions.
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