Question

Show that the difference of the squares of two integers cannot be less than 3.​

Answer 1

Step-by-step explanation:

let n and m be an integer and n>m and

m and n not equal to 0

then to prove :- n^2-m^2 > or = 3

since m is not equal to n

and n >m

then n^2-m^2= (n-m )(n+m)

here n-m not equal to 0

and n+m >n

so least possible value of n-m=1

and n+m = 2m+1

so 1×2m+1 =2m+1

so if we take least positive integer

m=1

then it is 2×1+1= 3

so 2m+1 > or equal to 3