a^2-b^4=2009
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Gaurav Seth 4 years, 3 months ago
a² - b⁴ = 2009
a² - b⁴ = (a + b²)(a - b²)
Now, we have to find out all the factors of 2009.
Factors of 2009 are = 1, 7, 41,49, 287 and 2009
Now, we have to use the difference of squares of factorization to obtain (a + b²)(a - b²) = 2009
The prime factorization of 2009 is 7²*41.
If we choose two factors 'x' and 'y' such that xy = 2009,
a + b² = x and a - b² = y, then 2b² = x - y.
If x = 2009, then y = 1 and 2b² = 2008,
then, b² = 2008/2
b = √1004 which is not and integer.
Now, if x = 287, then y = 7 and 2b² = 280
then, b² = 280/2
b = √140 which is also not an integer.
Now, if x = 49, then y = 41 and 2b² = 8
then b² = 8/2
b = √4
b = 2
Now, putting the value of b =2 in a² - b⁴ = 2009
⇒ a² - 2⁴ = 2009
⇒ a² - 16 = 2009
⇒ a² = 2009 + 16
⇒ a² = 2025
⇒ a = √2025
⇒ a = 45
So, the value of a and b is 45 and 2 respectively.
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