Question

Prove that x^n- a^n is divisible by x+a only when n is even.​

Answer 1

Step-by-step explanation:

I'm trying to prove that xn+yn is divisible by (x+y), when x and y are integers, and n is a positive, odd integer. I previously proved that xn-yn was divisible by (x-y) with induction, so I figured a similar method would make sense here, but I end up going in circles... this is what I tried:

Since n is an positive integer:

xn+yn can be rewritten as x2a+1+y2a+1, where a is a positive integer.

1) Base case (a=1)

x3+y3 = (x+y)(x2-xy+y2), and (x2-xy+y2) is an integer, so it is divisible.

2) Assumption (a=k)

Assume x2k+1+y2k+1 is divisible by (x+y)

3) Next 'step' (a=k+1)

x2(k+1)+1+y2(k+1)+1

=x2k+3+y2k+3

= (x2k+1+y2k+1)(x2) - x2y2k+1 + y2k+3

= (x2k+1+y2k+1)(x2) + (x2k+1+y2k+1)(y2) - x2k+1y2 - x2y2k+1

= (x2k+1+y2k+1)(x2) + (x2k+1+y2k+1)(y2) - (x+y)(x2ky2) + x2ky3 - x2y2k+1

= (x2k+1+y2k+1)(x2) + (x2k+1+y2k+1)(y2) - (x+y)(x2ky2) - (x+y)(y2kx2) + x3y2k +x2ky3 ... I tried a bunch more steps/other factors, but nothing seems to be working

Answer 2

Dear Student,

★ Answer:

It is Answered.

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★ Correction :

➠ Ques : Prove that $X^{n}$ - $a^{n}$  is divisible by  X + a only when X is even​

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★ Step-by-step explanation:

➼ The statement to be proved is:

P ( n ) :  $X^{n} - a^{n}$  is divisible by X + a when X is even​

➼ Step 1: Verify that the statement is true for the smallest value of X ,

here X =2

⇒ P ( 2 ) : $X^{2} - a^{2}$ is divisible by X + a

⇒ P (2) : (X + a) (X−a ) is divisible by X + a , which is true.

Therefore P(2) is true.

➼ Step 2 :  Assume that the statement is true for k

⇒ Let us assume that P (k) :  $X^{k} - a^{k}$  is divisible by X + a  where k is even.

➼ Step 3: Verify that the statement is true for the next possible integer, here for X =k+2

⇒ $X ^{k+2} - a^{k+2}$  =  $X^{k+2} - X^{2} a^{k} + X^{2} a^{k} - a^{k+2}$

= $X^{2} ( X^{k} -a^{k} ) + a^{k} (X^{2} - a^{2} )$

➥ Since,

⇒ $( X^{k} - a^k} )$  and $(X^{2} - a^{2} )$  are both divisible by (X + a ), the complete equality is divisible by X + a .

➥ Therefore,  

⇒ P ( k + 2)  :  $X ^{k+2} - a^{k+2}$ is divisible by X + a  where k+2 is even.

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Answer :

Therefore by principle of mathematical induction, P(n) is true.

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