find the pair of integer (a,b) such that a3+a2b+ab2+b3+1=2002 .
a3 + a2b + ab2 + b3+1 = 2002
⇒ (a+b) ( a2 + b2) = 2001 = 3 × 23 × 29
So, a+b is therefore one of the three numbers 3,23,29
Case 1:
When a+b = 3 then, a=1 ,b=2 or vice verse so a2 +b2 =12 +22 = 5.
Now, (a+b)(a2 +b2) = 3 × 5 = 15 ≠ 2001
Hence, this case is not possible.
Case 2:
When a+b = 23 and a2 + b2 = 87
Using hit and trial method, the least and maximum values are:
(1,22 ) and ( 11,12 )
so for (1, 22) we have, 12 + 222 = 485
⇒ (a+b) ( a2 + b2) = 23 × 485 ≠ 2001
Similarly, 112 + 122 = 121 + 144 = 265
⇒ (a+b) ( a2 + b2) = 23 × 265 ≠ 2001
Case 3:
When a + b =29 then a2 + b2 = 69
Again, using hit and trial method, the least and maximum values are:
(1,28 ) and (14 ,15)
12 + 282 >69
142 + 152 > 69 both of the taken values are not possible.
Therefore, all the three taken cases doesn't yield any solution.
Hence, the given pair of integers (a,b) have no solution.