You should prove that there exist no integer k such that (1+k^2) be divisible by 2 & 3 at the same time. 1+k^2 either is odd or even hence in each case is not divisible either by 2 or 3
If 6 time j is 1 more than the square of k, where k is an integer, what is the smallest possible value of j?
How can I show that it is indeed 1/6 ?
(Converting into symbols: 6j=k^2 + 1 but I'm stuck..)
If you use 0 as smallest integer , then k=0, so 6j = 1 : j = 1/6
If you use 1 as smallest integer , then k=1, so 6j = 2 : j = 1/3