Question

Smallest positive integer 'n' for which (2^2-1)(3^2-1)(4^2-1)...(n^2-1)is a perfect square is?​

Answer 1

Answer:

n = 8

Step-by-step explanation:

Given  

Smallest positive integer 'n' for which (2^2-1)(3^2-1)(4^2-1)...(n^2-1)is a perfect square is?

The given sequence can be written as 1 x 3, 2 x 4, 3 x 5 ------(n- 1)(n + 1)  

So 2n (n + 1) (3 x 4 x 5…….(n – 1))^2 should be a perfect square.

2n x (n + 1) must be a perfect square.

Now n and (n + 1) are co-prime numbers.  

So they must be perfect squares for 2n (n + 1) to be a perfect squares.

2n and n + 1 are perfect squares.

Now n and 2(n + 1) are perfect squares.

So n and (n + 1) is odd and other is even. Also 2 x odd is not a perfect square since power of 2 would be 1 in that case.

Hence odd of two number should be perfect square in itself and twice of even number should also be a perfect square.

Therefore, either of n or n + 1 must be a perfect square.

So either of n or n + 1 must be a perfect square. Now let n = odd^2 and odd^2 – 1 .

so n= 8 which makes a perfect square.

Answer 2

Answer:

n = 8

Step-by-step explanation:

Given  

Smallest positive integer 'n' for which (2^2-1)(3^2-1)(4^2-1)...(n^2-1)is a perfect square is?

The given sequence can be written as 1 x 3, 2 x 4, 3 x 5 ------(n- 1)(n + 1)  

So 2n (n + 1) (3 x 4 x 5…….(n – 1))^2 should be a perfect square.

2n x (n + 1) must be a perfect square.

Now n and (n + 1) are co-prime numbers.  

So they must be perfect squares for 2n (n + 1) to be a perfect squares.

2n and n + 1 are perfect squares.

Now n and 2(n + 1) are perfect squares.

So n and (n + 1) is odd and other is even. Also 2 x odd is not a perfect square since power of 2 would be 1 in that case.

Hence odd of two number should be perfect square in itself and twice of even number should also be a perfect square.

Therefore, either of n or n + 1 must be a perfect square.

So either of n or n + 1 must be a perfect square. Now let n = odd^2 and odd^2 – 1 .

so n= 8 which makes a perfect square.