prove for every odd integer a that a2 3a2 732b for some inte

prove for every odd integer a that (a2 +3)(a2 +7)=32b for some integer b

Solution

a is odd

let a =1

(a2 +3)(a2 +7)=4*8=32

compairing with 32b, b=1

let a=3

(a2 +3)(a2 +7)=12*16=192=32*6

compairing with 32b, b=6

let a= 2m+1

(a2 +3)(a2 +7)=[(2m+1)²+3][(2m+1)² +7]

=[4m² + 4m+4][4m² +4m+8]

=16[m² +m+1][m² +m+2]

[m² +m+1][m² +m+2] will always be multiple of 2

so, we can write [m² +m+1][m² +m+2]= 2b

hence (a2 +3)(a2 +7)=32b for every a when a is odd

proved