Using the formulae

S(p,q)= q*(p+q-1), and

D(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), and
D(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, and
D(1,5) = D(7,2) = D(9,1) = 5.

S(1,10) = S(13,4) + S(17,2), ie 100 = 64 + 36, and
D(1,10) = D(13,4) = D(17,2) = 10.

S(1,25) = S(15,18) + S(49,1), ie 625 = 576 + 49, and also
S(1,25) = S(31,10) + S(41,5), ie 625 = 400 + 225, and

D(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), etc

sums, which are squares for any values of n.

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?

Thankyou

sfwc
<><

1
3, 4
5, 8, 9
7, 12, 15, 16
9, 16, 21, 24, 25
11, 20, 27, 32, 35, 36

I will denote the number in row i, column j as <i, j>.
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 <i+1,1> - <i,1> = 2, those in column 2 by 4, ie <i+1,2>-<i,2> = 4, and generally,

<i+1,j> - <i,j> = 2j.

For all <i,j>, j<=i.

The value in <i,j> is 2*i*j - j^2. Therefore all <n,n> 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.

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 so
S(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 <i,j> values, it will be simply saying that if there is an <i,j> 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)^2
S(4*n+5, n^2) = n^2*(2n+1)^2
S(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 square
for 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, etc

You 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 <i, j> 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^2
4n+5, n^2
12n+7, 6n^2
8n+9, 2n^2
20n+11, 10n^2
12n+13, 3n^2
28n+15, 14n^2

Do you see the pattern? There are two sequences within it, running like this

Sequence 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, and

Sequence 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> i<>j squares.

If all this is true than we can generate Pythagorean triples!

>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> 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)

Thankyou

sfwc
<><

Thank you.

2n^2+2n+1, which gives us the sequence

5, 13, 25, 41, 61, 85, 113, 145, etc

The 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 general

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?

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 get

9, 25, 49, 81, 121 etc, which are parts of Pythagorean triples:

9+16=25
25+144=169
49+225=625
81+1600=1681
121+3600=3721.

Thankyou

sfwc
<><

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), 1<k<=n, that is, S1, S2 and S3 are all sums of consecutive odd integers, with the last integer in each sequence being the same a(n) for all three. Again easy: as S1 is a square, there is an integer j < n such that S1 = sum a(1)...a(j), and then
S2 = S3 - S1 = sum a(j+1) ... a(n), and therefore k = j + 1.

A similar construction is possible for cubes : if we add up consecutive elements of the sequence a(n) = 3n^2 - 3n + 1
(1, 7, 19, 37, 61, etc) starting with 1, we obtain cubes (1 = 1^3; 1 + 7 = 2^3; 1 + 7 + 19 = 3^3, etc).
If we could obtain a cube by adding up consecutive terms in a(n) = 3n^2 - 3n + 1, starting with a(i), i>1, 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

For 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?