## The Simple Proof of Beal's Conjecture

For the discussion of math. Duh.

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MrAwojobi
Posts: 38
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### Re: The Simple Proof of Beal's Conjecture

z4lis

all I can say is don't hate, appreciate. Read my proof with understanding and look for genuine loop holes and I bet you won't find any.

joshz
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### Re: The Simple Proof of Beal's Conjecture

Except your proof, as stated, does have a HUGE "loophole."
2 + 3 = 5, but 2 and 3 have no common factors. You can't expect us to infer what you mean if you say something totally different.

And since that false assumption doesn't hold, neither does your proof of Beal's Conjecture.
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Penitent87
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Location: London

### Re: The Simple Proof of Beal's Conjecture

MrAwojobi wrote:Penitent87

you are simply being obstinate. You keep substituting numbers into Beal's equation. The expression cannot be factorised without numbers.

That's not what I did at all. In fact, the numbers I substituted wouldn't be valid to sub into Beal's equation (I made sure of this). It seems you have a problem with the distiction between factorising something algebraically and a prime factorisation.

Remember that every positive number greater than one has a unique prime factorisation, but not every polynomial (or other function) can be factorised nicely in the way you are thinking. Does this mean that once numbers are substituted the result will have no prime factors? No.

Remember that you are trying to prove that if we have two products added together and they equal another number (which of course they must) then all three numbers must have a common factor. This is clearly not true, as shown by a number of posters. To show this, simply add together any two numbers with highest common factor 1. You clearly get another number which must have a prime factorisation and the original two numbers have no common factor. This contradicts what you are saying in your 'simple proof'.

I am aware I am probably wasting my time trying to convince you.

z4lis
Posts: 767
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### Re: The Simple Proof of Beal's Conjecture

MrAwojobi wrote:z4lis

all I can say is don't hate, appreciate. Read my proof with understanding and look for genuine loop holes and I bet you won't find any.

Here a few holes in your proof that I have found in the first two lines, underlined so they are easily spotted:

SIMPLE PROOF OF BEAL’S CONJECTURE
(THE \$100 000 PRIZE ANSWER)

What response can anyone have in the face of so many, many holes in a proof?
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MrAwojobi
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### Re: The Simple Proof of Beal's Conjecture

antonfire

I will give an example here by factorising 16 + 4 =20
4(4 + 1) = 20
16, 4 and 20 share common factors, 2 and 4

joshz
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### Re: The Simple Proof of Beal's Conjecture

5 + 2 = 7.
What common factors do any two of those share?

5 + 3 = 8.
What common factors do any two of those share?
Last edited by joshz on Tue Aug 10, 2010 12:00 am UTC, edited 1 time in total.
You, sir, name? wrote:If you have over 26 levels of nesting, you've got bigger problems ... than variable naming.
suffer-cait wrote:it might also be interesting to note here that i don't like 5 fingers. they feel too bulky.

TheGrammarBolshevik
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### Re: The Simple Proof of Beal's Conjecture

Hmm, ok, I'll bite:

MrAwojobi wrote:It isn’t difficult to see that the only way for

the 1st product + the 2nd product = the 3rd product

is if and only if the left hand side of the equation can be factorised.

It's difficult for me to see this. Can you explain why it is the case?
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Penitent87
Posts: 28
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Location: London

### Re: The Simple Proof of Beal's Conjecture

MrAwojobi wrote:antonfire

I will give an example here by factorising 16 + 4 =20
4(4 + 1) = 20
16, 4 and 20 share common factors, 2 and 4

Good for you.

But one example doesn't prove the general case at all. In fact, if this is the logic you are using, why did you even bother trying to prove the conjecture? Clearly it is true in one case, so it must hold in all cases...

What an insight.

antonfire
Posts: 1772
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### Re: The Simple Proof of Beal's Conjecture

MrAwojobi wrote:antonfire

I will give an example here by factorising 16 + 4 =20
4(4 + 1) = 20
16, 4 and 20 share common factors, 2 and 4
That is not an answer to my question.

What are you asserting here?
MrAwojobi wrote:It isn’t difficult to see that the only way for

the 1st product + the 2nd product = the 3rd product

is if and only if the left hand side of the equation can be factorised.

Are you suggesting that because the right hand side factors into a product of primes and the left hand side, A^x+B^y, does not factor symbolically, the two can never be equal?
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Talith
Proved the Goldbach Conjecture
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### Re: The Simple Proof of Beal's Conjecture

SIMPLE PROOF OF GOLDBACH'S CONJECTURE
(THE \$100 000 000 PRIZE ANSWER)
Goldbach’s Conjecture
Goldbach’s conjecture states that if n is an even integer greater than 2, then there exist primes p and q such that p + q =n

Examples

4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 7 + 3 or 5 + 5
12 = 5 + 7
14 = 3 + 11 or 7 + 7

Some Primes (cos that's what this is about, see)

A prime is one in where it has no prime factors (because it's prime itself lol) Examples are….

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173
179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281
283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409
419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541
547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659
661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797 809
811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941
947 953 967 971 977 983 991 997 1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069
1087 1091 1093 1097 1103 1109 1117 1123 1129 1151 1153 1163 1171 1181 1187 1193 1201 1213 1217 1223
1229 1231 1237 1249 1259 1277 1279 1283 1289 1291 1297 1301 1303 1307 1319 1321 1327 1361 1367 1373
1381 1399 1409 1423 1427 1429 1433 1439 1447 1451 1453 1459 1471 1481 1483 1487 1489 1493 1499 1511
1523 1531 1543 1549 1553 1559 1567 1571 1579 1583 1597 1601 1607 1609 1613 1619 1621 1627 1637 1657
1663 1667 1669 1693 1697 1699 1709 1721 1723 1733 1741 1747 1753 1759 1777 1783 1787 1789 1801 1811
1823 1831 1847 1861 1867 1871 1873 1877 1879 1889 1901 1907 1913 1931 1933 1949 1951 1973 1979 1987
1993 1997 1999 2003 2011 2017 2027 2029 2039 2053 2063 2069 2081 2083 2087 2089 2099 2111 2113 2129
2131 2137 2141 2143 2153 2161 2179 2203 2207 2213 2221 2237 2239 2243 2251 2267 2269 2273 2281 2287
2293 2297 2309 2311 2333 2339 2341 2347 2351 2357 2371 2377 2381 2383 2389 2393 2399 2411 2417 2423
2437 2441 2447 2459 2467 2473 2477 2503 2521 2531 2539 2543 2549 2551 2557 2579 2591 2593 2609 2617
2621 2633 2647 2657 2659 2663 2671 2677 2683 2687 2689 2693 2699 2707 2711 2713 2719 2729 2731 2741
2749 2753 2767 2777 2789 2791 2797 2801 2803 2819 2833 2837 2843 2851 2857 2861 2879 2887 2897 2903
2909 2917 2927 2939 2953 2957 2963 2969 2971 2999 3001 3011 3019 3023 3037 3041 3049 3061 3067 3079
3083 3089 3109 3119 3121 3137 3163 3167 3169 3181 3187 3191 3203 3209 3217 3221 3229 3251 3253 3257
3259 3271 3299 3301 3307 3313 3319 3323 3329 3331 3343 3347 3359 3361 3371 3373 3389 3391 3407 3413
3433 3449 3457 3461 3463 3467 3469 3491 3499 3511 3517 3527 3529 3533 3539 3541 3547 3557 3559 3571

Hence the reason why n has to be greater than 2 in Goldbach’s conjecture. For the purpose of the proof of Goldbach’s conjecture, the author wants to stress right from the onset that the primes of Goldbach are an entirely different type of prime to other primes even though they look similar. It should therefore not be surprising that it has to be exempt from Goldbach’s equation. The difference between these 2 equations is simply due to the fact that n = p + q can be rewritten as n = (p + 1) + (q - 1), i.e. n is a sum of 2 numbers not prime i.e. it can be summarised. n = p + q cannot be summarised hence the big difference between the 2 equations.

Simple Proof

It should be clear that each term in the equation n = p + q can be broken down into the sum of its primes. It isn’t difficult to see that the only way for

the first prime + the second prime = thebignumberwewanttomaketheprimesequalto

is if and only if the left hand side of the equation can be summarised. This will therefore guarantee that p and q add together to make n.

Penitent87
Posts: 28
Joined: Wed Apr 15, 2009 2:30 am UTC
Location: London

### Re: The Simple Proof of Beal's Conjecture

Talith wrote:SIMPLE PROOF OF GOLDBACH'S CONJECTURE
(THE \$100 000 000 PRIZE ANSWER)

It should be clear that each term in the equation n = p + q can be broken down into the sum of its primes. It isn’t difficult to see that the only way for

the first prime + the second prime = thebignumberwewanttomaketheprimesequalto

is if and only if the left hand side of the equation can be summarised. This will therefore guarantee that p and q add together to make n.

I can't believe noone thought of it before.

gmalivuk
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### Re: The Simple Proof of Beal's Conjecture

Sorry to ruin everyone's fun again.
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