Problem of the Day: June 17
For which positive integers m does the equation (ab)2015=(a2 + b2)m have positive integer solutions?
Edited to add solution: The combination of a product of a and b on one side and a sum of a and b on the other side immediately suggests two approaches: (i) AM-GM, and (ii) looking for common factors. We'll make use of both ideas.
Suppose p is a prime factor of a. Then p is also a factor of (ab)2015,and thus must be a factor of (a2 + b2)m. Thus p is also a factor of a2 + b2. Since p is a factor of a2, it must also be a factor of b2, and hence also a factor of b.
By symmetry, the same applies to b. Thus a and b have exactly the same prime factors. That's almost enough to conclude a=b (which would make the problem much easier), but not quite enough -- a and b might have the same prime factor but to different powers.
So suppose a=cpA and b=dpB where c and d are relatively prime to p, and B>A. Then we have p2015(A+B)(cd)2015 = p2AM(c2+pB-Ad2)m. Thus p2015(A+B)-2AM(cd)2015 = (c2+pB-Ad2)m. Since the right-hand side is not divisible by p, the left-hand side must not be, either, so we must have 2015(A+B)-2AM = 0.
But that can't happen. That's because m can't be very big. That's clear enough by inspection of the original equation -- since the power of the left-hand side is fixed, if m gets too big, there won't be integer solutions in a and b. By AM-GM, we know a2 + b2 >= 2ab > ab, so to have identity in our original equation, we must have m < 2015. But if m < 2015, and B > A, we can't have 2015(A+B) - 2AM = 0.
That then shows that a and b can't have different powers of p. That's enough to show that a = b. Now we go to work on the new simplified equation. We now have a4030 = (2a2)m, or a4030-2m= 2m. Thus
From this we see that a must be a power of 2, so let a=2k. Then 24030k-2mk = 2m. That gives 4030k-2mk=m. Thus 2mk - m + 4030k = 0. Now we use SFFT to get (m - 2015)(2k + 1) = -2015, or (2015-m)(2k+1)=2015. Running through factors of 2015, we get m =0, 1612, 1860, 1950, 1984, 2002, 2010, or 2014.
(Problem source: 2015 New Zealand Camp Selection Test #4)
Edited to add solution: The combination of a product of a and b on one side and a sum of a and b on the other side immediately suggests two approaches: (i) AM-GM, and (ii) looking for common factors. We'll make use of both ideas.
Suppose p is a prime factor of a. Then p is also a factor of (ab)2015,and thus must be a factor of (a2 + b2)m. Thus p is also a factor of a2 + b2. Since p is a factor of a2, it must also be a factor of b2, and hence also a factor of b.
By symmetry, the same applies to b. Thus a and b have exactly the same prime factors. That's almost enough to conclude a=b (which would make the problem much easier), but not quite enough -- a and b might have the same prime factor but to different powers.
So suppose a=cpA and b=dpB where c and d are relatively prime to p, and B>A. Then we have p2015(A+B)(cd)2015 = p2AM(c2+pB-Ad2)m. Thus p2015(A+B)-2AM(cd)2015 = (c2+pB-Ad2)m. Since the right-hand side is not divisible by p, the left-hand side must not be, either, so we must have 2015(A+B)-2AM = 0.
But that can't happen. That's because m can't be very big. That's clear enough by inspection of the original equation -- since the power of the left-hand side is fixed, if m gets too big, there won't be integer solutions in a and b. By AM-GM, we know a2 + b2 >= 2ab > ab, so to have identity in our original equation, we must have m < 2015. But if m < 2015, and B > A, we can't have 2015(A+B) - 2AM = 0.
That then shows that a and b can't have different powers of p. That's enough to show that a = b. Now we go to work on the new simplified equation. We now have a4030 = (2a2)m, or a4030-2m= 2m. Thus
From this we see that a must be a power of 2, so let a=2k. Then 24030k-2mk = 2m. That gives 4030k-2mk=m. Thus 2mk - m + 4030k = 0. Now we use SFFT to get (m - 2015)(2k + 1) = -2015, or (2015-m)(2k+1)=2015. Running through factors of 2015, we get m =0, 1612, 1860, 1950, 1984, 2002, 2010, or 2014.
(Problem source: 2015 New Zealand Camp Selection Test #4)
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