Bay 12 Games Forum

Some people have phobophobia. Replace everything with spiders. It won't help them though.

Well for lack of Middle East thread.

It is possible that the president of Iran, and some of his ministers are dead.
Their helicopter 'made a hard landing' near the Azerbaijan border.

The Iranian State television reports the helicopter has been found. The Iranian Red Half Moon denies this is the case and say search and rescue teams are still searching.

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At the upper end, a dumb storage of every single possible position[1] vector-multiplied with every possible ruleset[2] wouldn't technically need to be AIed

More than atoms in the universe. Good luck storing that.

I'm suggesting repurposing the place they store all the universes[/i[, obviously.

"Since when was the Church an Army?"
"Since most of your history. You've been living in a Blip."

I love that they went there.

Especially since Ewadd will not be directly providing us tribute or taxes, and given we very specifically refrained from looting the surrounding area, we should probably keep going so we can actually get some material compensation.

A

At the upper end, a dumb storage of every single possible position[1] vector-multiplied with every possible ruleset[2] wouldn't technically need to be AIed

More than atoms in the universe. Good luck storing that.

By that I mean, is there a natural number b such that a³ - b² = b+1 and (b+1)² - a³ = b, or alternatively, a³ - (b+1)² = b and b² - a³ = (b+1)? Let's call the first set of conditions H, and the alternative set of conditions G.

Very interesting question. I'd like to see the python code too, mine was rubbish and chugged the laptop hard when checking a up to 10,000.

Had a look at this but haven't made much progress on the why of it. I'm not a number theorist though. Here's a few things:

Facts about squares tell us that the first part of H is enough, i.e. a^3 = b^2 + b + 1 means that a^3 + b is (b+1)^2.

Wolfram gave me 4 integer solutions for a^3 - b^2 - b - 1 = 0 , but the other 3 rely on negative or 0 a or b.

I think you can just reformulate the problem as: 4 a^3 = k^2 +3 for some integers a,k, but I don't know why that only has the 4 integer solutions.

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You're right that G is impossible for positive integer a and b. There's a nice way of seeing why:

suppose G holds for some integer a, b

Then a^3 = b (b+1)

b and b+1 share no common factors... so must be cubes of integers themselves, as otherwise a would not be a cube.

The only integer cubes with a difference of 1 are -1, 0 and 1. So we obtain b=0 or b+1 = 0, so a=0 and b=-1 or b=0

Would love to see other people chime in on this, but I'd advise you to check if it's a case of an unsolved problem in number theory before getting too invested. There's a lot of those, and some can be stated very simply.

There was the sound of the very air tearing as Mildred forcibly slowed her charge with powerful counter winds, her missed attack leaving a furrow in the dirt. By the time she was reoriented, Akihiko had caught up and was already attacking back, hitting her with a kick that she couldn't fully block in time. Her forearm and shoulder took the majority of the blow and she let herself roll with it, cartwheeling wildly through the air and letting the momentum play itself out while rising up out of reach before stopping herself. She floated upside down in space for a moment, considering her plan, then corkscrewed around and moved in again.

As she dived down toward Akihiko, her body became alight with bluish-white fire, which she channeled into her fist for a powerful Superman-style punch.