Question

show that the difference of the square of two integers can never be lees than three

Answer 1

Answer:

Step-by-step explanation:

here is an example of 12

The sum of 12 consecutive integers is

n+(n+1)+(n+2)+⋯+(n+11)

which is equal to

12n+11×122=12n+66=6(2n+11)

For this number 6(2n+11) to be a perfect square, the number (2n+11) must either be equal to 6×12, or equal to 6×22, or equal to 6×32, etc. So, in general, we have 2n+11=6m2 for some integer m. This means that any satisfactory value for (2n+11) must be a multiple of 6. But this is not possible because (2n+11) is an odd number, since 2n is an even number, and any multiple of 6 is an even number. Thus, 6(2n+11) cannot be a square number.