Aha I see it. Any even one a sub 2k can be written in the form (a sub k)^2-100^k which neatly factors to (a sub k+10^k)(a sub k-10^k) for all evens. Sorry for my latex. And the odd case is trivial.
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Sal Gard
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6 months, 3 weeks ago

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@Sal Gard
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I have another unanswered discussion question. Can you take a look at it?
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D K
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6 months, 3 weeks ago

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@Sal Gard
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Nice observation. Kudos.
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D K
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6 months, 3 weeks ago

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That's interesting. I can show that if \(k\) is composite, then \(a_k\) is composite and has a factor of 101...101.
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Chung Kevin
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6 months, 3 weeks ago

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@Chung Kevin
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you mean if k is odd ? then ak will have 101 as a factor. but I also need proof for when k is even.
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D K
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6 months, 3 weeks ago

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TopNewestAha I see it. Any even one a sub 2k can be written in the form (a sub k)^2-100^k which neatly factors to (a sub k+10^k)(a sub k-10^k) for all evens. Sorry for my latex. And the odd case is trivial. – Sal Gard · 6 months, 3 weeks ago

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– D K · 6 months, 3 weeks ago

I have another unanswered discussion question. Can you take a look at it?Log in to reply

– D K · 6 months, 3 weeks ago

Nice observation. Kudos.Log in to reply

That's interesting. I can show that if \(k\) is composite, then \(a_k\) is composite and has a factor of 101...101. – Chung Kevin · 6 months, 3 weeks ago

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– D K · 6 months, 3 weeks ago

you mean if k is odd ? then ak will have 101 as a factor. but I also need proof for when k is even.Log in to reply