So it's one greater than a "near" square whose sides are a square and one more than that square.
1 + 4x5 = 3 x 7 1 + 9x10 = 7 x 13 1 + 16x17 = 13 x 21 1 + 25x26 = 21 x 31
Interesting! Is that pattern always true? 1 + (n^2)(n^2 + 1) =? (n^2 - (n-1))(n^2 + n + 1) = (n^2 - n + 1)(n^2 + n + 1) = n^4 + n^3 + n^2 - n^3 - n^2 - n + n^2 + n + 1 = n^4 + n^2 +1 which equals the original expression, which can therefore always be expressed as the product of at least two factors.
I saw this problem posted as an assertion that 10101 is not prime in any base. I started to take the same approach as you, but quickly realized that you can just factor n^4 + n^2 + 1 if you are sneaky enough.
I thought back to the derivation of the quadratic formula, and came up with this:
4 comments:
1 + n^2 + n^4 is always odd.
2: 21 no (3 x 7)
3: 91 no (7 x 13)
4: 273 no (3 x 91)
5: 651 no (3 x 217)
Hmm.
Here's another expression:
1 + n^2(1 + n^2)
So it's one greater than a "near" square whose sides are a square and one more than that square.
1 + 4x5 = 3 x 7
1 + 9x10 = 7 x 13
1 + 16x17 = 13 x 21
1 + 25x26 = 21 x 31
Interesting! Is that pattern always true?
1 + (n^2)(n^2 + 1) =? (n^2 - (n-1))(n^2 + n + 1)
= (n^2 - n + 1)(n^2 + n + 1)
= n^4 + n^3 + n^2 - n^3 - n^2 - n + n^2 + n + 1
= n^4 + n^2 +1
which equals the original expression, which can therefore always be expressed as the product of at least two factors.
I saw this problem posted as an assertion that 10101 is not prime in any base. I started to take the same approach as you, but quickly realized that you can just factor n^4 + n^2 + 1 if you are sneaky enough.
I thought back to the derivation of the quadratic formula, and came up with this:
n^4 + n^2 + 1 = n^4 + 2n^2 - n^2 + 1
= n^4 + 2n^2 + 1 - n^2
= (n^2 + 1)^2 - n^2 [Aha! a^2 - b^2!]
= (n^2 - n + 1)(n^2 + n + 1)
Factored, QED.
"Near" square numbers play a part in #100. My first effort took 19 hours to find a solution, and I've only managed to get that down to 38 minutes.
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