Question 184232
Let k=first whole number, (ie k is NOT a fraction or decimal number)


So k+1 would be the next consecutive whole number



Now square {{{k+1}}} to get {{{(k+1)^2=k^2+2k+1}}}


Now subtract {{{k^2}}} (the square of the first number) from {{{k^2+2k+1}}} (the square of the second) to get


{{{(k^2+2k+1)-k^2=(k^2-k^2)+2k+1=2k+1}}}



Now remember, we said at the top that "k" is a whole number. So this means that {{{2k+1}}} is guaranteed to be an odd number (since {{{2k}}} is an even integer)



So it is NOT possible to get an even difference between the squares of two consecutive whole numbers.



Examples (these don't prove the statement above, but help show it)


ex 1: Pick a number 6 and the next number 7. Square 6 to get 36. Square 7 to get 49


Subtract: 49-36=13


The difference is odd



ex 1: Select the number 12 and the next number 13. Square 12 to get 144. Square 13 to get 169


Subtract: 169-144=25


The difference is odd


You can try any two values and you'll find that the difference is odd.