Factorization is the decomposition of an expression into a product of its factors.
The following are common factorizations.
For any positive integer \(n\), \[a^n-b^n = (a-b)(a^{n-1} + a^{n-2} b + \ldots + ab^{n-2} + b^{n-1} ).\] In particular, for \( n=2\), we have \( a^2-b^2=(a-b)(a+b)\).
For an odd positive integer,
. .
.
Factorization often transforms an expression into a form that is more easily manipulated algebraically, that has easily recognizable solutions, and that gives rise to clearly defined relationships.
1. Find all ordered pairs of positive integer solutions such that .
Solution: We have . Since the factors and on the right hand side are integers whose product is a power of 2, both and must be powers of 2. Furthermore, their difference is
implying the factors must be and . This gives , and thus . Therefore, is the only solution.
2. Factorize the polynomial
Solution: Observe that if , then ; if , then ; and if , then . By the Remainder-Factor Theorem, and are factors of . This allows us to factorize
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For the worked example - 1 : there are 2 solutions : (3,3) and (3,-3)
(y+1)-(y-1) = 2
implies y + 1 = -2 or y + 1 = 4 and y -1 = -4 or y-1 = 2
Thus, the two solutions you have are (3,3) and (3,-3)
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Thanks. I added in "positive integers".
Thanks for this.
I did not understand why should (y+1) - (y-1) = 2 ?
Can anyone explain ?
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What do you not understand?
What do you think (y+1)−(y−1) is equal to ?
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I did not see it correctly . My fault ! Sir, I have a problem. I want to learn Number theory as is organized here on Brilliant but I don't follow anything beginning from modular inverses.I have tried the wikis but I still don't follow .Please suggest something.
How is their (y+1)-(y-1) difference 2?
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(y+1)−(y−1) =y+1−y+1 =1+1=2
(Y+1)-(y-1) = y+1-y+1= 2
2
How to think that both difference is 2. What is the main moto behind thinking such that????