Recognizing the Square Numbers!
( 2 n − 1 ) , ( 3 2 n − 1 ) , ( 4 3 n − 1 ) , ( 5 4 n − 1 ) , ( 6 5 n − 1 ) , ( 7 6 n − 1 ) , ( 8 7 n − 1 ) , ( 9 8 n − 1 )
In Calvin's number set, there are the numbers as listed above for each natural number n , and there are no other numbers in the set. How many square numbers does this set contain?
Note : The set of Natural Numbers doesn't contain the number 0 .
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2 solutions
@Satyajit Mohanty It is not universally agreed whether 0 ∈ N . You should've said "positive integer".
Alternatively,
Case 1 :
a ( a − 1 ) n is a square number.
Then one less than a square number is never a square number, ruling out case 1.
Case 2 :
a ( a − 1 ) n is not a square number.
Then either ( a − 1 ) n > 1 , implying that ( a − 1 ) n ≥ 3 , which violates FLT, or ((a-1)n = 1), which implies a = 2 and n = 1 . This is entirely possible, so we only have one possible square in the entire set.
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How does it violate FLT? a ( a − 1 ) n − 1 ( a − 1 ) n is a square, not necessarily a ( a − 1 ) n 'th power.
− 1 is not a quadratic residue mod 3 , 4 , 7 . We're left with 2 n − 1 for n = 1 and ( 5 2 n ) 2 − 1 cannot be a square, since the only squares apart by 1 are 0 , 1 .