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erszega
Member since Apr-23-05
Apr-25-05, 06:58 AM (EST)    "integer sequences and Pythagorean triples"

 All this is based on the simple observation that 1+3 is a square, 1+3+5 is a square, 1+3+5+7 is a square, etc.I use the following sequence:F(p,1) = p,F(p,q) = p + 2q -2,and so the sequence : p,p+2,p+4,...,p+2q-2.The sum of the sequence F(p,q) is S(p,q), where(1) S(p,q)= q*(p+q-1) Clearly, S(1,q) is always a square, because, using (1), S(1,q) = q^2.On the other hand, for example, if we choose p=4n+3, and q=2n^2, then S( 4n+3, 2n^2 ) = 4n^2*(n+1)^2, ie a square, for any n.The following sequence sums are all similar :S(4n+3, 2n^2) = 4n^2*(n+1)^2S(8n+3, 8n^2) = 16n^2*(2n+1)^2S(12n+3, 18n^2) = 36n^2*(3n+1)^2S(16n+3, 32n^2) = 64n^2*(4n+1)^2...S(4n+5, n^2) = n^2*(n+2)^2S(8n+5, 4n^2) = 4n^2*(2n+2)^2S(16n+5, 16n^2) = 16n^2*(4n+2)^2S(24n+5, 36n^2) = 36n^2*(6n+2)^2...S(12n+7, 6n^2) = 36n^2*(n+1)^2S(24n+7, 24n^2) = 144n^2*(2n+1)^2...S(8n+9, 2n^2) = 4n^2*(n+2)^2S(16n+9, 8n^2) = 16n^2*(2n+2)^2...S(20n+11, 10n^2) = 100n^2*(n+1)^2S(40n+11, 40n^2) = 400n^2*(2n+1)^2...etc.I call the above sequence sums P-sums, and the sequences behind them P-sequences, "P" standing for "Pythagorean".Please note that, for example,S(1,5) = S(7,2) + S(9,1), ie 25=16+9, using (1).If we introduce(2) D(p,q) = (p-1)/2+q, "D" possibly meaning "depth", although better Descriptions may be possible,then D(1,5) = D(7,2) = D(9,1) = 5Another example, with D=25, isS(1,25) = S(15,18) + S(49,1), ie 625 = 576 + 49, andS(1,25) = S(31,10) + S(41,5), ie 625 = 400 + 225, andD(1,25) = D(15,18) = D(49,1) = D(31,10) = D(41,5) = 25.Now I have the following four hypotheses, which I cannot prove:p1, q1, p2, q2, p3, q3 below denote integers.H1 : For any S(p1,q1)=S(p2,q2)+S(p3,q3), D(p1,q1)= D(p2,q2)= D(p3,q3), if there are integers k,l, and m, such that S(p1,q1)=k^2, S(p2,q2)=l^2, and S(p3,q3)=m^2.H2 : The value of p1 in S(p1,q1) in H1 is always 1.H3 : Any S(p,q) value, with p<>1, is generated by a P-sum.H4: For any S(p2,q2), where p2 <>1, there exist integers p1, p3, q3, such that S(p2,q2) + S(p3,q3) = S(1,q1).

Subject     Author     Message Date     ID integer sequences and Pythagorean triples erszega Apr-25-05 TOP RE: integer sequences and Pythagorean triples sfwc Apr-25-05 1 RE: integer sequences and Pythagorean triples erszega Apr-25-05 2 RE: integer sequences and Pythagorean triples erszega Apr-26-05 3 RE: integer sequences and Pythagorean triples sfwc Apr-26-05 4 RE: integer sequences and Pythagorean triples erszega Apr-26-05 5 RE: integer sequences and Pythagorean triples sfwc Apr-26-05 6 RE: integer sequences and Pythagorean triples alexb Apr-26-05 7 RE: integer sequences and Pythagorean triples erszega Apr-26-05 8 RE: integer sequences and Pythagorean triples Ramsey2879 May-22-05 12 RE: integer sequences and Pythagorean triples erszega Apr-28-05 9 RE: integer sequences and Pythagorean triples sfwc Apr-28-05 10 RE: integer sequences and Pythagorean triples erszega May-11-05 11 RE: integer sequences and Pythagorean triples Ramsey2879 May-26-05 13

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sfwc
Member since Jun-19-03
Apr-25-05, 11:24 AM (EST)    1. "RE: integer sequences and Pythagorean triples"
In response to message #0

 I must confess I have not fully understood everything you have said. However, I think I understand enough to prove or disprove some of your hypotheses. Please set me straight on anything I have misunderstood.>H1 : For any S(p1,q1)=S(p2,q2)+S(p3,q3), D(p1,q1)= >D(p2,q2)= D(p3,q3), if there are integers k,l, and m, such >that S(p1,q1)=k^2, S(p2,q2)=l^2, and S(p3,q3)=m^2. I believe this is false, citing the counterexample p1 = 25, q1 = 1, p2 = 7, q2 = 2, p3 = 1, q3 = 3, k = 5, l = 4, m = 3.>H2 : The value of p1 in S(p1,q1) in H1 is always 1. The same counterexample suffices.>H3 : Any S(p,q) value, with p<>1, is generated by a P-sum. I am unsure what you mean by this. Please define a P-sum more clearly.>H4: For any S(p2,q2), where p2 <>1, there exist integers >p1, p3, q3, such that S(p2,q2) + S(p3,q3) = S(1,q1). True. Suppose s = S(p2, q2) is odd. Then let p3 = 1, q3 = (s-1)/2, q1 = (s+1)/2. If s is even it must be divisible by 4 if it is square, so let p3 = 1, q3 = s/4 - 1 and q1 = s/4 + 1.What suggested these hypotheses to you?Thankyousfwc<>< erszega
Member since Apr-23-05
Apr-25-05, 01:14 PM (EST)    2. "RE: integer sequences and Pythagorean triples"
In response to message #1

 You are right. Sorry, I did jump to conclusions too quickly after recognising some patterns. erszega
Member since Apr-23-05
Apr-26-05, 09:01 AM (EST)    3. "RE: integer sequences and Pythagorean triples"
In response to message #2

 Let me try another hypothesis:Using the formulaeS(p,q)= q*(p+q-1), andD(p,q)=(p-1)/2+q,whenever we find an S(p1,q1) value which is a square, and p1 > 1, then we will find another S(p2,q2) square with p2 > 1, and also a q3, such that:S(p1,q1) + S(p2,q2) = S(1,q3), andD(p1,q1) = D(p2,q2) = D(1,q3)(p1, p2, q2, p3, q3 are integers).For example:S(1,5) = S(7,2) + S(9,1), ie 25=16+9, andD(1,5) = D(7,2) = D(9,1) = 5.S(1,10) = S(13,4) + S(17,2), ie 100 = 64 + 36, andD(1,10) = D(13,4) = D(17,2) = 10.S(1,25) = S(15,18) + S(49,1), ie 625 = 576 + 49, and alsoS(1,25) = S(31,10) + S(41,5), ie 625 = 400 + 225, andD(1,25) = D(15,18) = D(49,1) = D(31,10) = D(41,5) = 25.What I am trying to say is that the S(p,q) squares with p>1 are never "alone" for D(p,q), but are part of a Pythagorean triple. Can you prove or disprove?Later, I would like to say more about the S(4n+3,2n^2),S(4n+5,n^2),S(12n+7,6n^2), etcsums, which are squares for any values of n. sfwc
Member since Jun-19-03
Apr-26-05, 03:48 PM (EST)    4. "RE: integer sequences and Pythagorean triples"
In response to message #3

 >whenever we find an S(p1,q1) value which is a square, and p1 >> 1, then we will find another S(p2,q2) square with p2 > 1, >and also a q3, such that: >>S(p1,q1) + S(p2,q2) = S(1,q3), and >D(p1,q1) = D(p2,q2) = D(1,q3) >>(p1, p2, q2, p3, q3 are integers). Try p1 = 4, q1 = 1. Then S(p1, q1) = 2^2. Now D(p1, q1) = 5/2 = D(1, q3) = q3. So q3 would not be an integer. The same problem arises whenever p1 is even. However, if p1 is odd, let S(p1, q1) = k^2 and let p2 = 2k + 1, q2 = D(p1, q1) - k, q3 = D(p1, q1). Evidently we have D(p1, q1) = D(p2, q2) = D(1, q3). Now, S(1, q3) = D(p1, q1)^2 is evidently square. Also, S(q2, q3) = (D(p1, q1) - k)(D(p1, q1) + k) = D(p1, q1)^2 - k^2 = ((p1-1)/2 + q1)^2 - q1(p1 + q1 - 1) = ((p1-1)/2)^2 is square. Finally, S(p1, q1) + S(p2, q2) = k^2 + D(p1, q1)^2 - k^2 = D(p1, q1)^2 = S(1, q3).For example, in the case p1 = 7, q1 = 2, we have S(p1, q1) = 4^2 and D(p1, q1) = 5, so we let p2 = 2*4 + 1 = 9, q2 = 5 - 4 = 1 and q3 = 5, the exact numbers in your example.So your hypothesis is false for p1 even, true for p1 odd. How did you come up with it?Thankyousfwc<>< erszega
Member since Apr-23-05
Apr-26-05, 03:48 PM (EST)    5. "RE: integer sequences and Pythagorean triples"
In response to message #3

 I have now attached a file which demonstrates my point.It'shows a table with the following pattern: 1 3, 4 5, 8, 9 7, 12, 15, 16 9, 16, 21, 24, 2511, 20, 27, 32, 35, 36I will denote the number in row i, column j as .For instance,<1, 1> = 1<3, 2> = 8<4, 4> = 16 etc.You can notice that the numbers in column 1 grow by 2 with every step, ie - = 2, those in column 2 by 4, ie - = 4, and generally, - = 2j.For all , j<=i.The value in is 2*i*j - j^2. Therefore all values are squares.Let me go back to my earlier definitions of:S(p,q)= q*(p+q-1), andD(p,q)=(p-1)/2+q.D(p,q) represents a row in my table, and so (p-1)/2+q = i.q is the equivalent of a column, therefore q = j.To show that this works, we can look at an example:<4,2>=12 (by looking it up in the table, or using the 2*i*j-j^2 formula).Then q=j=2, and p=2*(i-q)+1=5, and soS(5,2)= 12 again, using the S(p,q)=q*(p+q-1) formula.My revised hypothesis (see my previous note) is that if S(p1,q1) is a square and p1>1, then there will be p2>1, and q2, p3 >0 such that S(p1,q1)+S(p2,q2)=S(1,q3), all three being integer squares, and D(p1,q1)=D(p2,q2)=D(1,q3).With my table with the values, it will be simply saying that if there is an with a square value in row i, and j<>i, then we will find at least one Pythagorean triple in that row, but there will never be a "lonely" square in any row, not forming a Pythagorean triple, apart from those with i=j.Again, this is a hypothesis, I cannot prove it, but I have not found a counterexample yet.I also referred to the following sums:S(4*n+3, 2*n^2) = 4n^2*(n+1)^2S(4*n+5, n^2) = n^2*(2n+1)^2S(12*n+7, 6*n^2) = 36*n^2*(n+1)^2,etc.,ie all these are squares.Now, if S(4*n+3, 2*n^2) is always a square, and q=j=2*n^2, and p=4*n+3, and so i = 2*n^2 + 2*n + 1 (using i = (p-1)/2+q), then <2*n^2+2*n+1, 2*n^2> identifies the squares in column 2*n^2 !eg, for n=1 <5,2>=16, ie a squarefor n=2, <13,8>=144, etc.Now, if my hypothesis is correct, then <2*n^2+2*n+1, 2*n^2) can be used to find Pythagorean triples, because then there will be a triple in row 2*n^2+2*n+1 for every n. To put it in another way, there will be at least three different S(p,q) values forming a triple, whenever D(p,q)=2*n^2+2*n+1. Interestingly, 2*n^2+2*n+1 will also generate the squares in column one. Simple to prove: if i = 2*n^2+2*n+1, and j = 1, then p = 2*(i-j)+1 = 4*n^2+4*n+2=(2*n+1)^2, ie a square, and P((2*n+1)^2,1)=(2*n+1)^2. Indeed, for n= 1,2,3,..., the 2*n^2+2*n+1 values are: 5, 13, 25, 41, 61, 85, 113, 145, etcYou can check: <5,1>, <13, 1>, <25, 1>, <41, 1>, <61, 1>, etc are all squares. Now something else in connection with the sequence S(4*n+3,2*n^2)= 4*n^2*(n+1)^2, S(4*n+5, n^2)= n^2*(n+2)^2, S(12*n+7, 6*n^2)= 36*n^2*(n+1)^2, etc., all of which generate squares, with identifiable coordinates : there strongly appears to be a pattern again, which I cannot prove, but this would be my second hypothesis: I list the p, q values for which I have checked that S(p,q) is a square (I will skip the '*' sign now):4n+3, 2n^24n+5, n^212n+7, 6n^28n+9, 2n^220n+11, 10n^212n+13, 3n^228n+15, 14n^2Do you see the pattern? There are two sequences within it, running like thisSequence 1 : 4n+3, 2n^2 ; (4+8)n+3+4, (2+4)n^2 ; (4+8+8)n+3+4+4, (2+4+4)n^2 ; (4+8+8+8)n+3+4+4+4, (2+4+4+4)n^2, andSequence 2 : 4n+5, n^2 ; (4+4)n+5+4, (1+1)n^2 ; (4+4+4)n+5+4+4, (1+1+1)n^2.So my second hypothesis is that these S(an+b,cn^2) sequences follow a pattern, and so can be generated. Then these sequences can be used to generate all the i<>j squares.If all this is true than we can generate Pythagorean triples! sfwc
Member since Jun-19-03
Apr-26-05, 05:57 PM (EST)    6. "RE: integer sequences and Pythagorean triples"
In response to message #5

 >My revised hypothesis (see my previous note) is that if >S(p1,q1) is a square and p1>1, then there will be p2>1, and >q2, p3 >0 such that >S(p1,q1)+S(p2,q2)=S(1,q3), all three being integer squares, >and >D(p1,q1)=D(p2,q2)=D(1,q3). >With my table with the values, it will be simply >saying that if there is an with a square value in row >i, and j<>i, then we will find at least one Pythagorean >triple in that row, but there will never be a "lonely" >square in any row, not forming a Pythagorean triple, apart >from those with i=j. >>Again, this is a hypothesis, I cannot prove it, but I have >not found a counterexample yet. Please note that since the nature of your construction implies that any p1 produced by this method will be odd, the hypothesis is true. See my previous post for a proof.>So my second hypothesis is that these S(an+b,cn^2) sequences >follow a pattern, and so can be generated. Then these >sequences can be used to generate all the i<>j >squares. Let us look closely at what we require here:S(an+b, cn^2) = cn^2(cn^2 + an + b - 1) is to be a square for all n.That is, (cn + a/2)^2 + bc - c - a^2/4 is to be a square for all n.So a necessary and sufficient condition is that a be even and (b-1)c = (a/2)^2. If you are willing to count this as a 'pattern' then your hypothesis is correct.>If all this is true than we can generate Pythagorean >triples! You may be interested to know that there is already a method in existence for generating pythagorean triples, which is somewhat simpler than the one you are suggesting. See for example https://www.cut-the-knot.org/pythagoras/pythTriple.shtml(Incidentally, on that page the line "As an aside, those who mastered the arithmetic of complex numbers might have noticed that (m + in)^2 = (n^2 - m^2) + i2mn" is incorrect; it'should be (n + im)^2)Thankyousfwc<>< alexb
Charter Member
1531 posts
Apr-26-05, 05:58 PM (EST)    7. "RE: integer sequences and Pythagorean triples"
In response to message #6

 >(Incidentally, on that page the line "As an aside, those who >mastered the arithmetic of complex numbers might have >noticed that (m + in)^2 = (n^2 - m^2) + i2mn" is incorrect; >it'should be (n + im)^2) Thank you. erszega
Member since Apr-23-05
Apr-26-05, 11:57 PM (EST)    8. "RE: integer sequences and Pythagorean triples"
In response to message #6

 Thank you. Ramsey2879 guest
May-22-05, 10:15 PM (EST)

12. "RE: integer sequences and Pythagorean triples"
In response to message #6

 >>My revised hypothesis (see my previous note) is that if >>S(p1,q1) is a square and p1>1, then there will be p2>1, and >>q2, p3 >0 such that >>S(p1,q1) S(p2,q2)=S(1,q3), all three being integer squares, >>and >>D(p1,q1)=D(p2,q2)=D(1,q3). >>With my table with the values, it will be simply >>saying that if there is an with a square value in row >>i, and j<>i, then we will find at least one Pythagorean >>triple in that row, but there will never be a "lonely" >>square in any row, not forming a Pythagorean triple, apart >>from those with i=j. Snip >>So my second hypothesis is that these S(an b,cn^2) sequences >>follow a pattern, and so can be generated. Then these >>sequences can be used to generate all the i<>j >>squares. >Let us look closely at what we require here: >S(an b, cn^2) = cn^2(cn^2 an b - 1) is to be a square >for all n. snip>So a necessary and sufficient condition is that a be even >and (b-1)c = (a/2)^2. Snip>>If all this is true than we can generate Pythagorean >>triples! >You may be interested to know that there is already a method >in existence for generating pythagorean triples, which is >somewhat simpler than the one you are suggesting. See for >example >https://www.cut-the-knot.org/pythagoras/pythTriple.shtml Snip>sfwc As I see it, a problem with the prior method is that it generates both primative and non-primative triples. I note that the proposed method will generate all possible Pythagorean triples, both primative and non-primative, but I believe it lends it'self to fine tuning such that only primative Pythagorean triples might be generated. The present conditions that a be even and (b-1)c = (a/2)^2 appear to me to be quite simple given that all Pythagorean triples are generated. I intend to look at this further. erszega
Member since Apr-23-05
Apr-28-05, 08:35 AM (EST)    9. "RE: integer sequences and Pythagorean triples"
In response to message #5

 In message #5, I mentioned the formula2n^2+2n+1, which gives us the sequence5, 13, 25, 41, 61, 85, 113, 145, etcThe differences between the consecutive numbers (ie F(n)-F(n-1) with F(i)=2i^2+2i+1) go like this:8, 12, 16, 20, 24, 28, 32, ...The formula for this, call it "difference sequence" is 4n+4.Can somebody please help and say how we can arrive from 4n+4 to 2n^2+2n+1, and, in generalhow can we find the formula F(n) for an integer sequence, the difference sequence of which is of the form an+b, where F(n)-F(n-1)=a(n-1)+b?As I can see, when the difference sequence is a(n)+a, then F(n) will be (a/2)n^2+(a/2)n+1, but why?Other:Just as an interesting point, to illustrate my main point in message #5:Multiplying each number in the 2n^2+2n+1 sequence by 2n-1, we get9, 25, 49, 81, 121 etc, which are parts of Pythagorean triples:9+16=2525+144=16949+225=62581+1600=1681121+3600=3721. sfwc
Member since Jun-19-03
Apr-28-05, 06:18 PM (EST)    10. "RE: integer sequences and Pythagorean triples"
In response to message #9

 >how can we find the formula F(n) for an integer sequence, >the difference sequence of which is of the form an+b, where >>F(n)-F(n-1)=a(n-1)+b? We want F(n) = cn^2 + dn + e.Substituting in gives:cn^2 + dn + e - c(n-1)^2 - d(n-1) - e = a(n-1) + b(a - 2c)(n-1) + b + d - c = 0So c = a/2, d = a/2 - b, and e can be anything.F(n) = a(n^2 + n)/2 - bn + e.Thankyousfwc<>< erszega
Member since Apr-23-05
May-11-05, 07:37 AM (EST)    11. "RE: integer sequences and Pythagorean triples"
In response to message #10

 Thank you.I have been thinking about this topic, and I think that the main idea is very simple:Taking the sequence of odd numbers 1,3,5,7,9, etc., defined by a(n)=2n-1, we can obtain squares by adding up the first n terms, starting with 1 (1=1^2, 1+3=2^2, 1+3+5=3^2, etc.). My point was the following: whenever we find i > 1, such that the sum of the consecutive terms in a(i),...,a(n), n>=i, is a square S1, then we will find another square S2 such that S1 + S2 = S3, and S3 = sum a(1)...a(n). Easy, we just have to choose S2 = sum a(1)...a(i-1). What my "hypothesis" also said was that S2 could then also be expressed as sum a(k)...a(n), 11, then S1 = sum a(1)...a(i-1) plus S2 = sum a(i)...a(n) would add up to S3 = sum a(1) ... a(n), S1, S2, and S3 being cubes. That would contradict Fermat's Last Theorem, therefore the following statement is true:Sum a(i) = 3i^2 - 3i + 1, i: from j to n, is a cube if and only if j=1. My question is if there is a direct and (preferably) simple proof for the above statement, without having to involve FLT.Kind regards Ramsey2879 guest
May-26-05, 08:19 AM (EST)

13. "RE: integer sequences and Pythagorean triples"
In response to message #11

 Erszega wrote>I have now attached a file which demonstrates my point. >It'shows a table with the following pattern: >> 1 > 3, 4 > 5, 8, 9 > 7, 12, 15, 16 > 9, 16, 21, 24, 25 >11, 20, 27, 32, 35, 36 >Snip>The value in is 2*i*j - j^2. Therefore all >values are squares. >>Let me go back to my earlier definitions of: >S(p,q)= q*(p+q-1), and >D(p,q)=(p-1)/2+q. >>D(p,q) represents a row in my table, and so >(p-1)/2+q = i. >q is the equivalent of a column, therefore q = j. >SnipThen p = 2i-2j+1 , q=j>It will be simply >saying that if there is an with a square value in row >i, and j<>i, then we will find at least one Pythagorean >triple in that row, but there will never be a "lonely" >square in any row, not forming a Pythagorean triple, apart >from those with i=j. >Snip>So my second hypothesis is that these S(an+b,cn^2) sequences >follow a pattern, and so can be generated. Then these >sequences can be used to generate all the i<>j >squares. >>If all this is true than we can generate Pythagorean >triples! I quoted the above from Erszega's post #5 as a means of setting a basis for my comments. In post #6 by sfwc, it was shown that for s(an+b,cn^2) to be square, a is even, and (b-1)c = a^2/4 are sufficient conditions.I will add the condition that b = odd to complete the requirements for Pythagorean triples and substitute t = a/2 such that (b-1)c = t^2.Pythagorean triple sets can be generated for any t under these conditions as long as we let C^2 = S(1,(2tn plus b-1)/2 Plus cn^2), A^2 be equal to S(2tn plus b,cn^2). B^2=C^2-A^2For instance if t=1, b=3 then c = 1/2 and the Pythagorean triples are (for n = 2m) A^2 = S(2n plus 3,n^2/2) and C^2= S(1,(n^2/2 plus n plus 1) and C^2-A^2 = B^2. These are all primitive triples since A=C-1.Any comments?

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