 Hello friends, welcome again to this session on algebraic identities and today we are going to take up one new special identity which is called Sophie Germain Identity, okay. Now this is a French mathematician Sophie Germain and she was in 18th century. So she was from 1776 you know to 1831. So this lady came up with this identity and the identity goes like this. So a to the power 4 plus 4 times b to the power 4 will be equal to a plus b whole square plus b square as one factor and a minus b whole square plus b square. Now this particular identity becomes very very useful if you have to prove that some expression is not a prime. Also if you have this kind of an expression then you can factor it in these two factors. Now let us see how it is derived. So a square a to the power 4 plus 4 times b to the power 4 can be written as a square square plus 2b square whole square is it like this. Now you can see this particular thing resembles some a square and this one resembles b square isn't it. So a square plus b square is there if I get 2ab term as well then I can complete the square and hence I can do something about it. So let us see how to go about it. So hence if you see a square whole square plus 2b square whole square and then we are adding 2a square times 2b square why because I have to complete the square and since I am adding this so I have to just you know subtract this item from the expression. Now the first three terms these three terms complete the square and it becomes a square plus 2b square whole square and this term is reduced to 4a square b square again if you see the whole expression is reduced to a square minus b square form isn't it. So hence we can deploy the identity a square minus b square is a plus b a minus b and hence you will get a square plus 2b square plus 2ab so plus 2ab and a square plus 2b square minus 2ab this is what the identity would look like. Now if you see again there is a square term of a binomial so if you see how 2b square can be split into b square and this b square isn't it. Now if you club them together a square plus b square plus 2ab you know already this is a plus b whole square and this b square comes here plus b square and similarly here also a square plus b square minus 2ab comes to a minus b whole square and this b square comes here okay so hence why did we learn we learned this particular identity called so feature main identity wherein fourth powers are factored and fourth factor fourth power especially of this type there has to be a four right so a to the power four plus four times b to the power four can be expressed like that isn't it now let us see an application of this identity so question is prove that n to the power four plus four is even is never a prime number is never a prime number for all n greater than one our n belongs to natural number set that is n is a natural number clearly if n equals to one so one plus four is five which is a prime number so hence this condition has been given so n has to be greater than one so how to solve so n to the power four plus four can be written as n to the power four plus four times one to the power four and now using Sophie German identity we can say let us say a is equal to n and b equal to one in that identity so n to the power four plus four to the power four into one to the power four can be expressed like this isn't it n plus one whole square plus one square times n minus one whole square plus one square so this is identity which I'm using I'm just substituting n for a and one for b right so hence this can be factored into two factors n plus one whole square plus one and n minus one whole square plus one right so there are two factors over and above one and the expression itself so anything which has more than two factors will never be a prime so whatever be the value of n you will always find n plus one whole square plus one and n minus one whole square plus one as two factors of this given expression isn't it so hence there are more than two factors and hence n to the power four plus four cannot be a prime number isn't it so using our Sophie German identity we could prove that this expression will never be a prime if n is greater than one