tag:blogger.com,1999:blog-12535639.post111554191454254632..comments2018-03-17T17:58:54.013-07:00Comments on Fermat's Last Theorem: Pythagorean Triples: SolutionLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger21125tag:blogger.com,1999:blog-12535639.post-7929451716575951702017-12-20T16:34:53.403-08:002017-12-20T16:34:53.403-08:00(a+b)^2-2ab=a^2+b^2
(3+4)^2-2*3*4=49-24=25
a^2+b^2...(a+b)^2-2ab=a^2+b^2<br />(3+4)^2-2*3*4=49-24=25<br />a^2+b^2=c^2?<br />c^2-a^2+b^2=0<br />second degree equation<br />c=(sqrt(4a^2+4b^2))/2<br />c=(sqrt(4*3^2+4*4^2)/2=5<br /><br />(y+1)^2=Y^2+2y+1<br />if 2y+1=x^2<br /> y+1= z<br />z^2=y^2+x^2<br />y=x^2-1/2<br />it seems that there is no demonstration for c^2<br />(y+1)^2-Y^2=2y+1<br /> z^2-y^2=2y+1<br />Nothing says that 2y+1 is a square number.Except by digital applications.Otherwise we must go back to Greek mathematicians.<br />ali bachouhttps://www.blogger.com/profile/03352395215751431528noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-23014594627682675642016-09-11T17:56:33.813-07:002016-09-11T17:56:33.813-07:00Good evening, my name is Renato.I'm degree in ...Good evening, my name is Renato.I'm degree in civil engineering. I found an interesting property about the primes. I wonder if the property is existing in the literature or not.<br /><br />p is a prime number if there is only one solution to the equation, x and y being integers. (The "trivial" solution is such that x + y = p and y = x + 1)<br /><br />The equation is:<br /><br />y² - x² = p<br /><br />Thank youRenato Ximeneshttps://www.blogger.com/profile/01725746092703155752noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-32207018367740739682016-01-20T06:34:48.277-08:002016-01-20T06:34:48.277-08:00can any one solve this problem to me.
Sketch the s...can any one solve this problem to me.<br />Sketch the surface model of a solid that simultaneously satisfies the following relation:<br />X2+y2≤Z2/4<br />Z ≥ 2, Z ≤ 9<br />Deepika........https://www.blogger.com/profile/05496385193629173924noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-65849995286233442152014-11-06T14:14:18.187-08:002014-11-06T14:14:18.187-08:00HI..., SHORT AND SIMPLE...HERE'S WHAT I CAME U...HI..., SHORT AND SIMPLE...HERE'S WHAT I CAME UP WITH:<br />x2+y2=z2<br />x=3<br />x2=9<br />y=4<br />y2=16<br />WHICH GIVES YOU:<br />z=5<br />z2=25Abdullah Abdullahhttps://www.blogger.com/profile/03787243390788445992noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-77170029901506685622012-03-21T03:33:17.104-07:002012-03-21T03:33:17.104-07:00greetings!
can you give me an algorithm or formul...greetings!<br /><br />can you give me an algorithm or formula that generate any pythagorean triple leg a and b wherein both legs could still be expressed as a sum of 2 squares ?<br /><br />regardsaprilhttps://www.blogger.com/profile/06910544328902427114noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-27203487673845163552011-05-29T01:20:40.191-07:002011-05-29T01:20:40.191-07:00There is an algorith to ind ALL Pythagorean triple...There is an algorith to ind ALL Pythagorean triples. I found it during my play around to find Fermat's original proof of his last theorem. For some reason, the pro's are not very interested, the silence is overwhelming. For those interested, I will be happy to publish it properly here, as long as its origin is never doubted. It will also show why Fermat's last theorem (the proof) is closely coupled to the same algoritm.olafhttps://www.blogger.com/profile/05290481527334392019noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-17237734524467042462010-08-24T22:40:26.484-07:002010-08-24T22:40:26.484-07:00Hi Larry,
For finding the pythago...Hi Larry,<br /> <br /> For finding the pythagorean triplet involving a given number i use this way. take the number square it and then factorize it in such a way that both factors are even or both odd.<br /><br />Ex: for 8<br />8^2 = 64 = 1*64 (1 odd ,1 even)<br /> 2*32 (2 even)<br /> 4*16 (2 even)<br /> <br />2*32 implies the other two numbers are (32+2)/2 and (32-2)/2<br /><br />4*16 implies (4+16)/2 and (16-4)/2<br /><br />so we can find all the triplets involving the given number.pichkarihttps://www.blogger.com/profile/07527436211308194953noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-75222522295662942262010-08-24T22:39:40.738-07:002010-08-24T22:39:40.738-07:00Hi Larry,
For finding the pythago...Hi Larry,<br /> <br /> For finding the pythagorean triplet involving a given number i use this way. take the number square it and then factorize it in such a way that both factors are even or both odd.<br /><br />Ex: for 8<br />8^2 = 64 = 1*64 (1 odd ,1 even)<br /> 2*32 (2 even)<br /> 4*16 (2 even)<br /> <br />2*32 implies the other two numbers are (32+2)/2 and (32-2)/2<br /><br />4*16 implies (4+16)/2 and (16-4)/2<br /><br />so we can find all the triplets involving the given number.pichkarihttps://www.blogger.com/profile/07527436211308194953noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-10894675248323367752008-08-11T15:12:00.000-07:002008-08-11T15:12:00.000-07:00This comment has been removed by the author.Evahttps://www.blogger.com/profile/08707956541954166680noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-2637030840280562392007-05-16T23:20:00.000-07:002007-05-16T23:20:00.000-07:00Thanks, Larry.You have answered my question.best r...Thanks, Larry.<BR/>You have answered my question.<BR/><BR/>best regards<BR/>cfChin Foohttps://www.blogger.com/profile/01083408961987703100noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-53112314787772209602007-05-16T00:03:00.000-07:002007-05-16T00:03:00.000-07:00Hi Chin Foo,If I understand your question correctl...Hi Chin Foo,<BR/><BR/>If I understand your question correctly, the answer is no.<BR/><BR/>While it is easy to find an all even solution by doubling any solution you would like, to get a primitive solution, all three integers need to be relatively prime. This means that at most, only 1 of the integers can be even.<BR/><BR/>Please let me know if I misunderstood your question.<BR/><BR/>I wasn't sure if this is what you are asking or if you are asking if x=6, how many different possible solutions are there.<BR/><BR/>Cheers,<BR/><BR/>-Larry<BR/><BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-73289893184017247122007-05-15T20:37:00.000-07:002007-05-15T20:37:00.000-07:00Good day, Larry.Sorry for not stated my question c...Good day, Larry.<BR/>Sorry for not stated my question clearly.<BR/><BR/>I never consider (6,8,10), (10,24,26) are original Pythagorean Triple since it is 2*(3,4,5) and 2* (5,12,13). <BR/><BR/>My question, is it possible to have a equation to find out Pythagorean Triple for x = 6 (but y not 8 and z not 10), x = 10 (but y not 24 and z not 26)?<BR/><BR/>Please advise.<BR/><BR/>thanks and best regards<BR/>cfChin Foohttps://www.blogger.com/profile/01083408961987703100noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-23822537142815847262007-05-14T21:38:00.000-07:002007-05-14T21:38:00.000-07:00Hi Chin Foo,I'm not clear on your question.A Pytha...Hi Chin Foo,<BR/><BR/>I'm not clear on your question.<BR/><BR/>A Pythagorean Triple that consists of all even numbers is 6,8,10 since<BR/><BR/>36 + 64 = 100<BR/><BR/>There are in fact an infinite number of them which can be derived from the formula in my blog.<BR/><BR/>Please let me know if I am misunderstanding your question.<BR/><BR/>Cheers,<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-88263889357841735172007-05-08T23:36:00.000-07:002007-05-08T23:36:00.000-07:00Hi, Larry.I discovered below equations to calculat...Hi, Larry.<BR/><BR/>I discovered below equations to calculate Pythagorean triples when I was a secondary school student.<BR/><BR/>For x = odd numbers. 1, 3, 5, 7,...<BR/>y = (x^2-1)/2, z = (x^2+1)/2<BR/><BR/>Eg.<BR/>1, 0, 1<BR/>3, 4, 5<BR/>5, 12, 13<BR/>7, 24, 25<BR/><BR/>For x = even numbers which could be divided by 4. 4, 8, 12, 16,...<BR/>y = (x^2/4)-1, z = (x^2/4)+1<BR/><BR/>Eg.<BR/>4, 3, 5<BR/>8, 15, 17<BR/>12, 35, 37<BR/>16, 63, 65<BR/><BR/>It is easy to proof that those above equations are correct for all intergers.<BR/><BR/>But until now I still can't find the phythegorean triple equations for all even numbers including, 2, 6, 10, 14....<BR/><BR/>Is it impossible?<BR/>Please advise.<BR/><BR/>thanks and best regards<BR/>cfChin Foohttps://www.blogger.com/profile/01083408961987703100noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1163491398070816882006-11-14T00:03:00.000-08:002006-11-14T00:03:00.000-08:00Hi Larry,Thank you very much.The question is right...Hi Larry,<BR/><BR/>Thank you very much.<BR/><BR/>The question is right. It should be no integer solution. Thanks again for show me the way to solve it.<BR/><BR/>Jennyjennyhttps://www.blogger.com/profile/08188638534844804212noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1163470088818548452006-11-13T18:08:00.000-08:002006-11-13T18:08:00.000-08:00Hi Jenny,Are you sure that you have stated the pro...Hi Jenny,<BR/><BR/>Are you sure that you have stated the problem correctly.<BR/><BR/>There are no integer solutions to the problem x^2 + y^2 = 7*z^2.<BR/><BR/>The reason for this is that any prime of the form 4n+3 that divides the sum of two pairs must itself have an even power in the prime factorization.<BR/><BR/>Unfortunately, it is impossible for 7=(4*1+3) to have an even power in the prime factorization 7*z^2.<BR/><BR/>Here is a <A HREF="http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture21/lecture21/node2.html" REL="nofollow">web page</A> that provides the proof.<BR/><BR/>Here is <A HREF="http://rutherglen.ics.mq.edu.au/math334s106/m2334.Dioph.squares.pdf" REL="nofollow">another proof</A> in pdf format.<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1163442652497972612006-11-13T10:30:00.000-08:002006-11-13T10:30:00.000-08:00Hi Larry,Could you please teach me how to solve th...Hi Larry,<BR/><BR/>Could you please teach me how to solve the below Diophantine equation?<BR/><BR/>Find all solutions to the Diophantine equation x^2 + y^2 = 7z^2. <BR/><BR/>Thank you very much.<BR/><BR/>Jennyjennyhttps://www.blogger.com/profile/08188638534844804212noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1142203313167324792006-03-12T14:41:00.000-08:002006-03-12T14:41:00.000-08:00Thanks for your post. Yes, that also works since:...Thanks for your post. <BR/><BR/>Yes, that also works since:<BR/><BR/>Let p be any positive integer.<BR/>Let x = 2p+1 (so, x is odd)<BR/>So, x^2 = (2p+1)^2 = 4p^2+4p+1<BR/><BR/>So, y = (4p^2 + 4p + 1 - 1)/2 = 2p^2 + 2p<BR/><BR/>So, z = (4p^2 + 4p + 1 + 1)/2 = 2p^2 + 2p + 1<BR/><BR/>Now, z^2 = (2p^2 + 2p + 1)^2 = 4p^4 + 8p^3 + 8p^2 + 4p + 1<BR/><BR/>And y^2 + x^2 = (2p^2 + 2p)^2 + 4p^2 + 4p + 1 = 4p^4 + 8p^3 + 8p^2 + 4p + 1.<BR/><BR/>Cheers,<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1141878248597540202006-03-08T20:24:00.000-08:002006-03-08T20:24:00.000-08:00Does this make sense?Does this make sense?kkrazhttps://www.blogger.com/profile/03770539037262414116noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1141877946135127032006-03-08T20:19:00.000-08:002006-03-08T20:19:00.000-08:00I've been thinking about this for a while and have...I've been thinking about this for a while and have a simple way to identify whole number pythagorean triples. Consider any odd number x.<BR/><BR/>Then, y = (x2 - 1)/2<BR/><BR/>and z = (x2 + 1)/2<BR/><BR/>For example if x = 5 (x2 = 25)<BR/>then y = 12 and z = 13.<BR/><BR/>For x = 7 (x2 = 49) <BR/>y = 24 and z = 25.<BR/><BR/>This works because x2 = y + z<BR/>and y + 1 = z<BR/><BR/>So x2 + y2 <BR/>= (y + z) + y2 <BR/>= z + (y + y2) <BR/>= z + (y * z) <BR/>= z2<BR/><BR/>For example, <BR/>if x = 5<BR/>25 + (12 * 12) = (13 * 13)<BR/>(12 + 13) + (12 * 12) = (13 * 13)<BR/>13 + (13 * 12) = (13 * 13)<BR/>(13 * 13) = (13 * 13)<BR/><BR/>So, basically, x2 + y2 = z2 works with whole numbers because there is <BR/>a number, which when squared is the difference between n2 and (n-1)2.kkrazhttps://www.blogger.com/profile/03770539037262414116noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1141877002363788902006-03-08T20:03:00.000-08:002006-03-08T20:03:00.000-08:00Larry - I am definitely a math amateur, but have b...Larry - <BR/><BR/>I am definitely a math amateur, but have been thinking about this for a while and have a couple of questions.kkrazhttps://www.blogger.com/profile/03770539037262414116noreply@blogger.com