Find two positive consecutive odd integers such that the square of the larger integer is one less than twice the square of the small integer.

Let
x---------------> first  positive odd integer
x+2------------> second  positive consecutive odd integer

we know that
(x+2)²=2x²-1---------> x²+4x+4=2x²-1-------> x²-4x-5
x²-4x-5=0
using a graph tool--------> to calculate the quadratic equation
see the attached figure

the solution is
x=5

the answer is
the first  positive odd integer x is 5
the second  positive consecutive odd integer x+2 is 7