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#51 2024-04-18 21:07:07
- AnthonyRBrown
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
It's all a Mystery ? Like trying to get blood out of a stone!
The Ruddest in the classroom...just might be the Genius! 
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#52 2024-04-23 01:16:06
- phrontister
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
It's all a Mystery ? Like trying to get blood out of a stone!
Indeed! 
Absolute accuracy is required for all digits (right up to and including the units digit)
The rounding that occurs once QB64's double-precision limit is reached has enabled me to manufacture the following two X³ + Y³ = Z³ 'solutions' (I adapted your FLT code):
Just BASIC (a scaled-down freeware version of Liberty BASIC) can handle HUGE numbers accurately, making 'solution' manufacturing rather difficult!
Here are images of JB's output for the same X, Y & Z input as for the above QB64 images:
So, no JB solutions there.
Btw, the yellow highlighted portions denote groups of similar digits.
Last edited by phrontister (2024-04-23 01:25:12)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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#53 2024-04-23 05:13:50
- AnthonyRBrown
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
It never gives up! 
The Ruddest in the classroom...just might be the Genius!
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#54 2024-04-23 11:43:34
- phrontister
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
It never gives up!
But wait...there's more!
The most important numbers are the end numbers in brackets () the reason they are the most important is because they are the end of the process for making cube numbers! they are if you like the TRUTH numbers! its a bit like the numbers go round and round mathematically,and eventually make a whole cube number!
Ok. So, given QB64's (mis)handling of numbers that are too large for it, I thought I'd post an example demonstrating the inaccuracies that can occur in QB64 coding with the all-important last-digit TRUTH numbers in cases where the 16-digit double-precision accuracy boundary is breached:
The Program A segment of that image shows QB64's accurate output of the last digit of X & Y values, compared with the inaccurate output of the last digit of Z values, each of which overstep the double-precision boundary.
In Program B of that image, all X,Y,Z values overstep the double-precision boundary, hence all those inaccuracies.
There are no such inaccuracies in the Just BASIC image below, because those same X,Y,Z values that I used in QB64 (see the image above) are within JB's HUGE upper precision limit...whatever that is! 
Last edited by phrontister (2024-04-23 13:33:16)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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#55 2024-04-23 15:47:14
- AnthonyRBrown
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
2024-04-10 11:18:46
AnthonyRBrown
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The FLT Demonstration program I made was made FREE! on a budget! It satisfies to show the Three Patterns A,B,C for the accuracy of the program up to certain level.
Most people read all the threads,and then walk away!
The Ruddest in the classroom...just might be the Genius!
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#56 2024-04-23 21:12:27
- phrontister
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
Most people read all the threads,and then walk away!
But I'm still here!
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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#57 2024-04-23 22:45:50
- AnthonyRBrown
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
Truth will out.
The Ruddest in the classroom...just might be the Genius!
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#58 2024-04-24 09:40:21
- phrontister
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
Truth will out.
Nae doot, mon! 
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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#59 2024-04-27 00:27:53
- phrontister
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
The rounding that occurs once QB64's double-precision limit is reached has enabled me to manufacture the following two X³ + Y³ = Z³ 'solutions'...
I've explored that further, and found that manufacturing X³ + Y³ = Z³ 'solutions' for ARB's QB64 program is fairly easily done...like so:
1. X¹ = a 17-digit number.
2. Y¹ = X¹.
Note: {Y¹ = X¹+1}, {Y¹ = X¹+2} and {Y¹ = X¹+3} also work well, but yield successively fewer 'solutions' than {Y¹ = X¹}.
3. Z¹ = the cube root of Z³: just select the first 17 digits of the cube root (the last digit may need rounding up if it's ≥5).
Using ARB's latest code (post #33), I added an option to enter custom X¹,Y¹,Z¹ numbers instead of using the program's random numbers. Here's my adapted code:
I ran that code a few minutes ago, and here's an image from it:
In my adapted code I restricted the length of the random numbers so that their cubes display accurately, with no rounding or truncation: ie, within QB64 double-precision limits.
Last edited by phrontister (Yesterday 09:31:57)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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#60 2024-04-27 00:40:52
- Bob
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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#61 Yesterday 09:33:45
- phrontister
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
phrontister (post #52) wrote:The rounding that occurs once QB64's double-precision limit is reached has enabled me to manufacture the following two X³ + Y³ = Z³ 'solutions'...
I've explored that further, and found that manufacturing X³ + Y³ = Z³ 'solutions' for ARB's QB64 program is fairly easily done...like so:
1. X¹ = a 17-digit number.
2. Y¹ = X¹.
Note: {Y¹ = X¹+1}, {Y¹ = X¹+2} and {Y¹ = X¹+3} also work well, but yield successively fewer 'solutions' than {Y¹ = X¹}.
3. Z¹ = the cube root of Z³: just select the first 17 digits of the cube root (the last digit may need rounding up if it's ≥5).
Inspired by the second image in my post #52, my theory was that for both X¹ & Y¹ >17 digits, there'd be 'solutions' if their lengths are equal and their first 17 digits are equal (or nearly equal). However:
1. My first trial, with X¹ & Y¹ both 24 digits long, resulted in no solutions until their first 19 digits were equal...got the "WOW!" output then.
2. My next trial, with X¹ & Y¹ both 30 digits long, resulted in no solutions (tried equal digits from first 15 to first 30, all in vain). Hmm...
3. Conclusion: There are many, many 'solutions', but I haven't found the 'rule' that discovers them all.
I'll quit my test there, I think.
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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#62 Yesterday 22:07:11
- AnthonyRBrown
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Re: Fermats Last Theorem Demonstration! by, Anthony.R.Brown (1998 Solved )
Proving Fermat' s Last Theorem (almost) in just 2 minutes !
https://www.youtube.com/watch?v=EymVXkPWxyc
The Ruddest in the classroom...just might be the Genius!
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