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One, two, three, five. (Posted on 2014-06-07) Difficulty: 3 of 5
Find all solutions for integer x,y, such that 5x+1=2*3y.

Prove that there are no others.

  Submitted by broll    
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Solution: (Hide)
Write as 5^x=2*3^y-1

Checking small values, 2*3^y-1 = 2, mod3, while 5^x alternates between 1 and 2, mod3.

Since 5^(2n)=1, mod3, we can tighten LHS to 5^(2x-1).

Since 5 does not divide RHS unless y={1,5,9,13..} RHS is of form 2*3^(4n-3)-1.

So now 5^(2x-1)=2*3^(4y-3)-1.

Since with the exception of 5 itself, 2*3^(4n-3)-1 = 8, mod9, RHS must = 8 mod 9, which is the case only if 5 is raised to a power of (6x-3):

2*3^(4n-3)-1 = 5^(6x-3)

Now LHS = {5,2,4}, mod 7, while RHS = 6, mod 7.

So {x=0,y=0} and {x=1,y=1} are the sole solutions.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: Hintbroll2014-06-15 04:15:07
Hints/TipsHintbroll2014-06-13 02:14:06
Some Thoughtsanswers, no proofAdy TZIDON2014-06-07 10:34:18
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