Prove that if n^2 + 1 is a prime number greater than 5, then the digit in the 1's place of n is 0, 4, or 6.?

Its easy to prove this :

If n^2 + 1 is a prime then n^2 is an even number => n is an even number.

If n ends with 2 or 8 then n^2 + 1 ends with 5 which cannot be a prime number since all numbers that end with 5 are divisible by 5.

So n must end with 0,4 or 6.

Let the digit in the 1’s place of n be represented by x.

If x=0 , then the digit in the 1’s place n^2+1 will be 0^2+1=1

If x=1 , then the digit in the 1’s place n^2+1 will be 2

If x=2 , then the digit in the 1’s place n^2+1 will be 5

If x=3 , then the digit in the 1’s place n^2+1 will be 0

If x=4 , then the digit in the 1’s place n^2+1 will be 7

If x=5 , then the digit in the 1’s place n^2+1 will be 6

If x=6 , then the digit in the 1’s place n^2+1 will be 7

If x=7 , then the digit in the 1’s place n^2+1 will be 0

If x=8 , then the digit in the 1’s place n^2+1 will be 5

If x=9 , then the digit in the 1’s place n^2+1 will be 2

n^2+1 will be a prime only if its digit in 1’s place is 7 or 1.

Because if digit in 1’s place is 2,6 or 0 then it is divisible by 2.And if it is 5 then it will be divisible by 5.

Thus from the above list it is clear that the only possible chance to become n^2+1 prime is that x (digit in 1’s place of n ) should be 0,4 or 6

n^2 can't be odd (ie end in 1, 3, 5, 7 or 9) because then adding 1 makes n^2 + 1 an even number

Since the last place in the number must be even, we just need to look at the lowest digit.

it can't be 1, 3, 5, 7 or 9, since these give 1, 9, 5, 9, 1 as the last digit

0 * 0 ==> 0

2 * 2 ==> 4

4 * 4 ==> 6

6 * 6 ==> 6

8 * 8 ==> 4

But 4 + 1 = 5, so, 2 and 8 are ruled out

So, we are left with 0, 4 and 6

n = 2k+1, since all primes > 5 are odd. n^4 - 1 = (2k)^4 + 4(2k)^3 + 6(2k)^2 + 4(2k) 16k^4 + 32k^3 + 24k^2 + 8k = S 240 = 16 * 3 * 5 If k is even, S is divisble by 2 * 8. If k is odd, 24 (k^2) + 8 k = 8 k (3k+1), and 3k+1 is even, so S is divisible by 2 * 8. Therefore S is divisible by 16. k mod 3 = 0 => S divisible by 3 k mod 3 = 1 => (2k+1) = 0 mod 3 => n not prime k mod 3 = 2 => (2k+1) = 2 mod 3, 2^4 = 1 mod 3, so n^4 - 1 is divisible by 3. n mod 5 = 0 => n not prime n mod 5 = 1 => (n^4 - 1) = 0 mod 5 n mod 5 = 2 => (n^4 - 1) = 0 mod 5 (16 -1 = 15) n mod 5 = 3 => (n^4 - 1) = 0 mod 5 (81 -1 = 80) n mod 5 = 4 => (n^4 - 1) = 0 mod 5 (256 - 1 = 255) Therefore n^4 - 1 divisible by 16, by 3, and by 5, so by 240. (OK, the other guy's proof is a little neater, but it's nice to see two different approaches to the problem.)

0^2 = 0 + 1. = 1. prime number, but NOT greater than 5; 4^2 = 16 + 1 = 17. prime number & greater than 5; 6^2 = 36 + 1 = 37. prime number & greater than 5. (:

Vffgg.