EXPONENTIAL DIOPHANTINE EQUATIONS PART1

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how to solve 4^x+5^y=6^z
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You can see that X=0 by odd/even parity.

z cannot be 0 as the min value of 6^z is 2. When z=1 we have
1 + 5^y = 6 so (0, 1, 1) is a solution.

If z>1 then 6^z == 0 (mod 4)
So 1 + 5^y == 0 (mod 4)
But for all y, 5^y == 1 (mod 4).
Therefore no more solutions.

mcwulf
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Just looking at the equation I instinctively say: x=0, y=1, z=1. 1+5=6, done!

SebastienPatriote
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You stole my theme track
"Hey big smoke, see this nigga here!!"

rahulvyas
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It depends on possibilities of base figures.

lalitupadhyay
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You did not prove that it's the only solution.

mcwulf