Question

Prove that difference between squares of two consecutive positive integers is always of the form
(2m+1), for some integer m.​

Answer 1

Let the smaller Integers be x so the other will be x+1.

Now,

${x}^{2} + {(x + 1)}^{2} \\ = {x}^{2} + {x}^{2} + 2x + 1 \\ = 2x {}^{2} + 2x + 1 \\ = 2( {x}^{2} + x) + 1$


Now if we make
${x}^{2} + x = m$

Then we have the sum as,
$2m + 1$

Hence proved.