Prove this...?

Question

Let p be an odd prime other than 5, Prove that exactly one of
(p^2)+1 and (p^2)-1 is divisible by 10

prove it for all odd primes other than 5
not just 7

Seriously
prove it for all odd primes other than 5
not finite numbers
i know that it works for 3,5,7,11,13,17...
I want to know if you can prove it for all primes (which is infinite) by using variables to represent the numbers

Answer 1

The trick to this one is:

This works for all odds other than multiples of 5 for the following reason:
p must end in 1, 3, 7 or 9 (since it can't end in 5, or it would be divisible by 5).
So any number ending in one of these digits can be expressed as 10t + 1, or 10t + 3, or 10t + 7, or 10t +9, where t is any positive integer. (This means 't' is the tens and higher digits.)

Square these expressions to get:
(10t + 1)^2 = 100t^2 + 20t + 1, which is one more than a multiple of 10
(10t + 3)^2 = 100t^2 + 60t + 9, which is one less than a multiple of 10
(10t + 7)^2 = 100t^2 + 140t + 49, which is one less than a multiple of 10
(10t + 9)^2 = 100t^2 + 180t + 81, which is one more than a multiple of 10.

Any questions?

Answer 2

Any odd number (it doesn't even have to be prime for this to be true!) that does not not end in 5 must end in 1, 3, 7, or 9, so when you square it, the square has to end in either 1 or 9 since 1^2 = 1, 3^2 = 9, 7^2 = 49, and 9^2 = 81. (That is, all numbers that end with 1 or 9 have a square that ends with 1, and all numbers that end with 3 or 7 have a square that ends with 9.) For the squares that end in 9, add 1 and you have a multiple of 10. For the squares that end in 1, subtract 1 and you have a multiple of 10.

Answer 3

7^2 + 1 = 50 11^2 - 1 = 120


cant use 9 its prime

Answer 4

are you sure this is the right info? cause using a number like 7, then 49 plus 1 equals 50, which is divisible by 10, but 49 minus 1 equals 48, which is not divisible. That's a weird problem.

Answer 5

Let 7 take the place of p.
49+1 = 50/10 = 5
Therefore, it is divisible
49-1 = 48/10 = 4.8
Therefore, it is not divisible by 10.
Thus, (p^2)+1 is divisible by 10 while (p^2)-1 is not.

Answer 6

let p=3 therefore (3^2)+1=10, which is divisible by 10
let p=7 therefore (7^2)+1=50, which is divisible by 10
let p=11 therefore (11^2)-1=120, which is divisible by 10
let p=13 therefore (13^2)+1=170, which is divisible by 10
let p=17 therefore (17^2)+1=290, which is divisible by 10
let p=19 therefore (19^2)-1=360, which is divisible by 10
Yo can try others number

Answer 7

go call up sum of ur friends and ask them if they get it