Why prime factors?

Number Theory Level pending

The number which is four more than the square of 625 has exactly two prime factors. Compute them.

3 and 7 cannot be determined 677 and 7 577 and 677

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3 solutions

Arulx Z
Dec 27, 2015

Sophie Germain identity can be used to factor the number.

Identity states that -

a 4 + 4 b 4 = ( a 2 + 2 b 2 + 2 a b ) ( a 2 + 2 b 2 2 a b ) { a }^{ 4 }+4{ b }^{ 4 }=\left( { a }^{ 2 }+{ 2b }^{ 2 }+2ab \right) \left( { a }^{ 2 }+{ 2b }^{ 2 }-2ab \right)

We can rewrite our equation as

= 625 2 + 4 = 5 8 + 4 = ( 5 2 ) 4 + 4 1 4 ={ 625 }^{ 2 }+4\\ ={ 5 }^{ 8 }+4\\ ={ \left( { 5 }^{ 2 } \right) }^{ 4 }+4\cdot { 1 }^{ 4 }

On factoring, the expression simplifies as 577 677 577\cdot 677 .

Moderator note:

Great usage of Sophie Germain identity :)

Nice approach

tanay gaurav - 5 years, 5 months ago
Azzim Habibie
Dec 21, 2015

625^2 + 4 = 625^2 + 2^2 = ( 625 + 2 )^2 - 2.2.625 = 627^2 - 50^2 = ( 627 - 50 )( 627 + 50 ) = 577 . 677

Tanay Gaurav
Dec 12, 2015

Number = 625^{2} + 4 = 25^{4} Let us consider, x^{4} + 4= x^{4} + 4.[x^{2}] + 4 - 4.[x^{2}] = [x^{2} + 2]^{2} -[2x]^4 = [x^{2} + 2 + 2x][x^{2} + 2 - 2x] Similarly, 5^{4} = ([5^{2}]^{2} + 2[5^{2}] + 2)([5^{2}]^{2} - 2[5^{2}] + 2) = 577.677 So the prime factors are 577 and 677.

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