Thread: 7- table
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Old 2008-07-25, 09:35   #36
Raman
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"Mr. Tuch"
Dec 2007
Chennai, India

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Quote:
Originally Posted by xilman View Post
Solve the equation 1025x = 1 (mod 1050) in integers. The solution is the integer you want.
Be careful! There exist no solution to this equation. Since 1025 is even, a multiple of it is always even, and on the right hand side, 1 (mod 1050) is always odd. A solution is impossible to exist!

Quote:
Originally Posted by xilman View Post
Substitute x = y + 1/y in the octic and see what you get ...
No hopes for degree 4. Substituting x = y + (1/y) in x8, so it gives up
\sum_{z=0}^8 ^8C_z y^z (1/y)^{8-z}
which is clearly being at degree 8.

Other terms will have their appropriate degrees. So, when substituted, the whole algebraic polynomial will be of degree 8 only.

And the linear polynomial becomes more cumbersome, in this form, with
1025(y+(1/y)) - (1050+1)
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