Question

Prove that the sequence 7,11,15.. doesn't contain any perfect square

Answer 1

Answer:

Step-by-step explanation:

This is what i can get after few attempts....  -->

The sequence 7,11,15,....   can be denoted as (4n+3) where n lies from 1 to inf

Now suppose a is square of some number which lies in this range So,

$a^2=4n+3\\a^2-(4n+3)=0$

Now we will solve for its roots

By  method of roots =  (-b+-root(D))/2a  where D is discriminant

b=0

a=1

$D=b^2-4ac\\D=0-4*1*-(4n+3)\\D=4(4n+3)$

Roots = +-4(4n+3)/2*1 = +- 2(2n+3)

So we found value of a = 2(2n+3)

Since it is an even number as it is multiple of 2

It's square will also be even number

But in our sequence (4n+3) will always lead to an odd number

So it can't lie in given sequence which disapprove our assumption

So the given sequence can't contain any perfect square

Hope it helps :-)