Show that the square of any odd integer is of the form (8k+1)

Any odd integer can be represented by 2m+1 for some integer m (can prove that 2m+1 is always an odd number if needed). (2m+1)²=4m²+2m+2m+1=4m²+4m+1=4m(m+1)+1 simple factorisation. m and (m+1) are consecutive integers, one of them has to be an even number, when an odd number is multiplied by an even number the result is always an even number (I will prove it to the student if needed), since m(m+1) is even it can be written in the form 2k for some integer k, now 4m(m+1)+1=4(2k)+1=8k+1 hence the square of any odd integer can be written in the form (8k+1). proof completed.