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.

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