Find the pairs...
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- Birottam Dutta
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a^3 + b^3 + ba^2 + ab^2 + 1= 2002
=> a^2 (a+ b) + b^2 (a+ b) = 2001
=> (a+b) (a^2+ b^2) = 2001
Now, 2001 = 3*23*29
As a product of two factors, 2001 can be expressed as 29*69 or 23*87 or 3*667.
Take 2001= 29*69. If these are to be expressed as (a+b)(a^2+ b^2), a+b=29 and (a^2 + b^2)= 69. On, checking it can be easily found that 69 cannot be expressed as a sum of squares of two integers. So, this combination is not possible.
Similarly, for 2001= 23 * 87, 87 also cannot be expressed as the sum of two different squares.
For 2001 = 3*667, a+ b =3, => a=1 and b=2 or vice versa. This will never give a^2 + b^2 = 667. Therefore, none of the equations hold true.
Thus, no pair of integers satisfies this equation.
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=> a^2 (a+ b) + b^2 (a+ b) = 2001
=> (a+b) (a^2+ b^2) = 2001
Now, 2001 = 3*23*29
As a product of two factors, 2001 can be expressed as 29*69 or 23*87 or 3*667.
Take 2001= 29*69. If these are to be expressed as (a+b)(a^2+ b^2), a+b=29 and (a^2 + b^2)= 69. On, checking it can be easily found that 69 cannot be expressed as a sum of squares of two integers. So, this combination is not possible.
Similarly, for 2001= 23 * 87, 87 also cannot be expressed as the sum of two different squares.
For 2001 = 3*667, a+ b =3, => a=1 and b=2 or vice versa. This will never give a^2 + b^2 = 667. Therefore, none of the equations hold true.
Thus, no pair of integers satisfies this equation.
-------------
If you feel like a thanks, click it
![Smile :)](./images/smilies/smile.png)