 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Further Arithmetic Integers (Posted on 2006-06-27) Consider three positive integers x < y < z in arithmetic sequence, and determine all possible solutions of: x4 + y4 = z4 - 64

 See The Solution Submitted by K Sengupta Rating: 2.5000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 3 of 10 | `If x = y-k and z = y+k, then`
`  x^4 + y^4 = z^4 - 64 becomes`
`  y*(8*k^3 + 8*k*y^2 - y^3) = 64`
`Therefore, y = 1, 2, 4, 8, 16, 32, or 64.`
`Of these, only three are possible:`
`  y = 2    :    5 = k*(k^2 + 4),  k = 1 or 5`
`  y = 4    :   10 = k*(k^2 + 16), k = 1, 2, 5, or 10`
`  y = 8    :   65 = k*(k^2 + 64), k = 1, 5, 13, or 65`
`Of these, only`
`  y = 2  and  k = 1`
`  y = 8  and  k = 1`
`are valid. Therefore,`
`  x < y < z -----------`
`  1 < 2 < 3`
`  7 < 8 < 9`
` `

 Posted by Bractals on 2006-06-27 21:01:09 Please log in:

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