Question

- Be prime numbers p and t as P = (2x +1) and t = (2Y +1), where x and y are integers,
- To show that whatever n, greater than or equal to 2, there are numbers x and y as x = (p-1) / 2 and y = (t-1) / 2, so that the equation n = x + y +1 is always true
- For n = 2 we have p = 2 and t = 2 so that 2 = (2-1) / 2 + (2-1) / 2 +1 ie 2 = 1/2 +1 / 2 +1 or 2 = 1 + 1 or 2 = 2
- For n = 3 we have p = 3 and t = 3 so that 3 = (3-1) / 2 + (3-1) / 2 +1 or 3 = 1 + 1 +1 ie
3 = 3
- For. n = 4 we have p = 5 and t = 3 so that 4 = (5-1) / 2 + (3-1) / 2 +1 or 4 = 2 + 1 + 1 ie
4 = 4
...
- For. n = k we have

Answer 1

- Be prime numbers p and t as P = (2x +1) and t = (2Y +1), where x and y are integers,
- To show that whatever n, greater than or equal to 2, there are numbers x and y as x = (p-1) / 2 and y = (t-1) / 2, so that the equation n = x + y +1 is always true
- For n = 2 we have p = 2 and t = 2 so that 2 = (2-1) / 2 + (2-1) / 2 +1 ie 2 = 1/2 +1 / 2 +1 or 2 = 1 + 1 or 2 = 2
- For n = 3 we have p = 3 and t = 3 so that 3 = (3-1) / 2 + (3-1) / 2 +1 or 3 = 1 + 1 +1 ie
3 = 3
- For. n = 4 we have p = 5 and t = 3 so that 4 = (5-1) / 2 + (3-1) / 2 +1 or 4 = 2 + 1 + 1 ie
4 = 4
...
- For. n = k we have k = x + y + 1 ie k = (p-1) / 2 + (t-1) / 2 +1 suppose is always true, so
- For. k = k + 1 we have k +1 = ((p-1) / 2 + (t-1) / 2 +1) +1 ie k + 1 = k + 1
so q.e.d.

- so can being accepted right,correct this proof ?