- #1

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- Homework Statement
- If ## p\neq5 ## is an odd prime, prove that either ## p^{2}-1 ## or ## p^{2}+1 ## is divisible by 10.

- Relevant Equations
- None.

Proof:

Suppose ## p\neq5 ## is an odd prime.

Applying the Division Algorithm produces:

## p=10q+r ## where ## 0\leq r\leq 10 ##,

where there exist unique integers ## q ## and ## r ##.

Note that ## p\equiv 1, 3, 7 ## or ## 9 ## mod ## 10 ##.

This means ## p^{2}\equiv 1, -1, -1 ## or ## 1 ## mod ## 10 ##.

Then we have ## p=10q+1, p=10q+3, p=10q+7 ## or ## p=10q+9 ##.

Now we consider four cases.

Case #1: Suppose ## p=10q+1 ## for some ## q\in\mathbb{Z} ##.

Then we have ## p^{2}-1=(10q+1)^{2} -1

=100q^{2} +20q+1-1

=100q^{2}+20q

=10(10q^{2}+2q)

=10m ##,

where ## m=10q^{2} +2q ## is an integer.

Thus, ## p^{2} -1 ## is divisible by ## 10 ##.

Case #2: Suppose ## p=10q+3 ## for some ## q\in\mathbb{Z} ##.

Then we have ## p^{2} +1=(10q+3)^{2} +1

=100q^2 +60q+10

=10(10q^{2} +6q+1)

=10n ##,

where ## n=10q^{2} +6q+1 ## is an integer.

Thus, ## p^{2} +1 ## is divisible by ## 10 ##.

Case #3: Suppose ## p=10q+7 ## for some ## q\in\mathbb{Z} ##.

Then we have ## p^{2} +1=(10q+7)^{2} +1

=100q^{2} +140q+49+1

=100q^{2} +140q+50

=10(10q^{2} +14q+5)

=10r ##,

where ## r=10q^{2} +14q+5 ## is an integer.

Thus, ## p^{2} +1 ## is divisible by ## 10 ##.

Case #4: Suppose ## p=10q+9 ## for some ## q\in\mathbb{Z} ##.

Then we have ## p^{2} -1=(10q+9)^{2} -1

=100q^{2} +180q+81-1

=100q^{2} +180q+80

=10(10q^{2} +18q+8)

=10s ##,

where ## s=10q^{2} +18q+8 ## is an integer.

Thus, ## p^{2} -1 ## is divisible by ## 10 ##.

Therefore, if ## p\neq 5 ## is an odd prime, then either ## p^{2} -1 ## or ## p^{2} +1 ## is divisible by ## 10 ##.

Suppose ## p\neq5 ## is an odd prime.

Applying the Division Algorithm produces:

## p=10q+r ## where ## 0\leq r\leq 10 ##,

where there exist unique integers ## q ## and ## r ##.

Note that ## p\equiv 1, 3, 7 ## or ## 9 ## mod ## 10 ##.

This means ## p^{2}\equiv 1, -1, -1 ## or ## 1 ## mod ## 10 ##.

Then we have ## p=10q+1, p=10q+3, p=10q+7 ## or ## p=10q+9 ##.

Now we consider four cases.

Case #1: Suppose ## p=10q+1 ## for some ## q\in\mathbb{Z} ##.

Then we have ## p^{2}-1=(10q+1)^{2} -1

=100q^{2} +20q+1-1

=100q^{2}+20q

=10(10q^{2}+2q)

=10m ##,

where ## m=10q^{2} +2q ## is an integer.

Thus, ## p^{2} -1 ## is divisible by ## 10 ##.

Case #2: Suppose ## p=10q+3 ## for some ## q\in\mathbb{Z} ##.

Then we have ## p^{2} +1=(10q+3)^{2} +1

=100q^2 +60q+10

=10(10q^{2} +6q+1)

=10n ##,

where ## n=10q^{2} +6q+1 ## is an integer.

Thus, ## p^{2} +1 ## is divisible by ## 10 ##.

Case #3: Suppose ## p=10q+7 ## for some ## q\in\mathbb{Z} ##.

Then we have ## p^{2} +1=(10q+7)^{2} +1

=100q^{2} +140q+49+1

=100q^{2} +140q+50

=10(10q^{2} +14q+5)

=10r ##,

where ## r=10q^{2} +14q+5 ## is an integer.

Thus, ## p^{2} +1 ## is divisible by ## 10 ##.

Case #4: Suppose ## p=10q+9 ## for some ## q\in\mathbb{Z} ##.

Then we have ## p^{2} -1=(10q+9)^{2} -1

=100q^{2} +180q+81-1

=100q^{2} +180q+80

=10(10q^{2} +18q+8)

=10s ##,

where ## s=10q^{2} +18q+8 ## is an integer.

Thus, ## p^{2} -1 ## is divisible by ## 10 ##.

Therefore, if ## p\neq 5 ## is an odd prime, then either ## p^{2} -1 ## or ## p^{2} +1 ## is divisible by ## 10 ##.