(y" - b") is completely divisible by (y - b), when
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Answer:
Step 1: putting n=1, we get
x2n−y2n=x2−y2=(x−y)(x+y)
above number is divisible by (x+y). Hence the statement is true n=1
step 2: assuming that for n=m, (x2n−y2n) is divisible by (x+y) then we have
(x2m−y2m)=k(x+y)
Where, k is an integer
step 3: putting n=m+1 we get
x2(m+1)−y2(m+1)
=x2m+2−y2m+2
=x2m+2−x2my2−y2m+2+x2my2
=(x2m+2−x2my2)+(x2my2−y2m+2)
=x2m(x2−y2)+y2(x2m−y2m)
Setting the value of (x2m−y2m) from (1), we get
x2m(x−y)(x+y)+y2k(x+y)
=k(x2m(x−y)+y2)(x+y)
It is clear that the above number is divisible by (x+y). Hence the statement holds for n=m+1
Thus (x2n−y2n) is divisible by (x+y) for all positive integers n≥1
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answered
Jul 29 '15 at 12:56
Harish Chandra Rajpoot
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Jul 29 '15 at 13:29
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8
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Hint: Rewrite:
x2k+2−y2k+2=(x2k+2−y2kx2)+(y2kx2−y2k+2)=x2(x2k−y2k)+y2k(x2−y2).
Added: Note it can be proved without induction:
x2n−y2n=(x2)n−(y2)n=(x2−y2)(x2(n−1)+x2(n−2)y2+⋯+x2y2(n−2)+y2(n−1)),
and the first factor is divisible by x+y.
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answered
Jul 29 '15 at 12:38
Bernard
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Jul 29 '15 at 13:23
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3
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x2(k+1)−y2(k+1)=x2x2k−y2y2k=x2x2k−x2y2k+x2y2k−y2y2k=x2(x2k−y2k)+(x2−y2)y2k.
The first term is divisible by x+y by the induction hypothesis, and the second as well, by direct factorization.
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answered
Jul 29 '15 at 12:47
Yves Daoust
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why must we have that green part? – momokjaaaaa Mar 26 '17 at 10:05
Because it allows the next step. – Yves Daoust Mar 27 '17 at 6:54
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For n=k, assume P(k) is true, we have
x2k−y2k=A(x−y)
, where A is a polynomial.
For n=k+1,
x2k+2−yk+2 =x2[A(x−y)+y2k]−y2k+2 =A(x−y)x2+x2y2k−y2k+2 =A(x−y)x2+y2k(x2−y2) =A(x−y)x2+y2k(x−y)(x+y) =(x−y)[Ax2+y2k(x+y)] =B(x−y), where B is a polynomial.
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answered
Jul 29 '15 at 12:39