Almost Pythagorean
Are there positive integer solutions to
x 3 + y 4 = z 5 ?
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1 solution
At first I thought this was a problem on using Fermat's Last Theorem or divisibility. I then toyed with the idea of scaling each variable by some constant, but that didn't work. Finally I tried x 3 = y 4 which would double the LHS, and since the RHS is a power of z it would be nice for the LHS to be a power as well, so I let ( x , y , z ) = ( 2 a , 2 b , 2 c )
Since x 3 = y 4 therefore a = 4 m , b = 3 m and 2 1 2 m + 1 = 2 5 c which meant that m was 2 m o d 5 .
This is sufficient to prove that there are infinitely many solutions, though I would love to see other solutions (or a proof of their nonexistence).
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How do u know it will only be power of 2 ?
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I don't know whether all solutions are powers of 2 . I wanted to show that there were infinitely many solutions of a specific form - you can think of it as a special case. The motivation was that the LHS would be doubled if x 3 = y 4 and I wanted to make both sides a power of 2 to compare indices. Not all solutions have to be of that form.
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@Shaun Leong – Yeah Infinitely many type of infinite sets of solutions exist. Yours is one of them. More can be found using Pythagorean Triplets.
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@Kushagra Sahni – Can you help add some examples of what you're thinking of?
It's not immediately apparent to me how we get these powers. AFAIK, this is the main family of solutions that people find.
How did you obtain this triplet? Is this the only parametrization?
Solutions of the form ( x , y , z ) = ( 2 4 ( 5 k + 2 ) , 2 3 ( 5 k + 2 ) , 2 1 2 k + 5 ) for a non-negative integer k satisfy the equation, and there are infinitely many of them.