Suppose that p is prime and p n2 5 for some natural number

Suppose that p is prime and p = n^2 + 5 for some natural number n. Prove that the final digit of p is equal to 1 or 9. That is, prove that p 1( mod 10) or p 9( mod 10).

Solution

For any natural number n, the last digit of n^2 will be one among 0,1,4,9 and 6.

When the last digit of n^2 is 0:

Then the last digit of n^2+5 is 5.

Then definitely p=n^2+5 is not prime.

So the last digit of n^2 cannot be zero.

When the last digit of n^2 is 1:

Then the last digit of n^2+5 is 6.

Then definitely p=n^2+5 is not prime.

So the last digit of n^2 cannot be 1.

When the last digit of n^2 is 4:

Then the last digit of n^2+5 is 9, which can be prime.

So the last digit of p can be 9.

When the last digit of n^2 is 9:

Then the last digit of n^2+5 is 4. (because 9+5=14)

Then definitely p=n^2+5 is not prime.

So the last digit of n^2 cannot be 9.

When the last digit of n^2 is 6:

Then the last digit of n^2+5 is 1.(Because 6+5=11)

Then n^2+5 can be prime.

So last digit opf p=n^2+5 can be 1.

So by above results, the last digit pf p can be either 1 or 9.