Is it a perfect square?
Determine the number of integers for which is a perfect square.
(This problem is a modified form of the one that appeared in KVS JMO 2014)
The answer is 0.
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n 6 + 3 n 5 − 5 n 4 − 1 5 n 3 + 4 n 2 + 1 2 n
= n 5 ( n + 3 ) − 5 n 3 ( n + 3 ) + 4 n ( n + 3 )
= ( n 5 − 5 n 3 + 4 n ) ( n + 3 )
= n ( n + 3 ) ( n 4 − 5 n 2 + 4 )
= n ( n + 3 ) ( n 2 − 4 ) ( n 2 − 1 )
= ( n − 2 ) ( n − 1 ) n ( n + 1 ) ( n + 2 ) ( n + 3 )
= Product of 6 consecutive integers
= A multiple of 6!
= Unit digit is 0
Unit digit of n 6 + 3 n 5 − 5 n 4 − 1 5 n 3 + 4 n 2 + 1 2 n + 3 is 3.
We see that unit digit of a square can be one of 0,1,4,9,6.
Thus , 0