Yes, it will enrich the literature. We may consider it for publication if it is a new result. 

If something like 32 + 42 + 122  = 132 . Nothing new as 32+ 42 = 52 and 52  + 122 = 132. a set of two Pythagorean numbers. Kindly give some detail.

Hi,

I hope that you found this link.

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html

The sums of consecutive cubes is ALWAYS equal to a perfect square and other interrelationships can also be derived. I am currently writing a paper on the relationship between squares and cubes in relation to the solution of the cubic diophantine equation x^3+y^3+z^3=a63

Sorry! Obviously I meant x^3+y^3+z^3=a^3

"The sums of consecutive cubes is ALWAYS equal to a perfect square"

What about 3^3+4^3+5^3 = 216 and √216 = 14.696938457?

You must start with 1^3. 1^3=1^2. 1^3+2^3=3^2. 1^3+2^3+3^3=6^2. 1^3+....+4^3=10^2. 1^3+...+5^3=15^2,etc The squares themselves form a sequence 1/3/6/10/15/21/28. The equation 3^3+4^3+5^3=6^3 is the classic cubic diophantine equation which I partially solved in my paper published in 2006 in Missouri Journal of Mathematical Sciences. A follow-up paper is now almost ready for submission for publication.

"You must start with 1^3"

I see. That follows from the closed expressions for formulas of type "sum for t=1 to n of t^k". For k=3, the closed form is 1/4(n^2)(n+1)^2.

Another way of looking at this is that the sum of consecutive cubes beginning with 1^3 equals the square of the sum of the integers being cubed. For example, 1^3+2^3+3^3+....+10^3=(1+2+3+....10)^2. Other interesting relationships can also be derived. The trick is to try to relate these relationships to the solution of the cubic diophantine equation x^3+y^3+z^3=a^3.  That is the thrust of my research.. Incidentally, 3^3+4^3+5^3=6^3 must be rewritten as 4^3+5^3+3^3=6^3 (See my 2006 publication). 

Louis these results are very basic

Probably true!  But then, I figured this all out by myself. I'm not a mathematician; I'm a doctor. However, take a look at my 2006 paper; that was a good piece of work. My current paper that is almost ready for submission is not as ground breaking but does offer further insight into the cubic diophantine equation. I'd be pleased to hear any comments that anyone has re the 2006 paper. Thanks.

In that case, I must congratulate you for showing your interest in mathematics. But you need to understand that trail and error method is not always leads you to the destiny. It is the technique that you need to develop so that one can apply it in some other situation too. 

Check the book by Grosswald  on Sums of squares and compare!  

Thank you for all your comments. My 2006 paper produced two solutions to the cubic diophantine equation. The first involved trial and error but the second was a mathematically rigorous derivation of a formulaic solution which had never been previously achieved and resulted in the paper being published as the lead article in that issue of the journal. My current work tries to relate properties of squares and cubes to this fascinating equation but still does not allow us to incorporate into the solution those equations where x is not a perfect square - presumably because such a solution does not exist. If anyone has any further insight into how one could do so, I'd love to hear about it. It's great to see this interest in something that has fascinated me for years,

Dear friends:

   Consider two pythagorean triples (x, y, z) and (z, Y, Z), with z repeated. For example, (3,4,5) and (5,12,13). Then, if (x,y,z) and (z,Y,Z) ---in this order---are written in a growing order, we have. x squared + ysquared = z squared; z squared + Y squared = Z squared. From these we get: x squared + y squared + Y squared = Z squared. in the above example,   3 squared + 4 squared + 12 squared = 13 squared.

   You see that this is a very special example, but there are infinitely many of them.

   (The above exercise is part of a paper which I haven't been able to upload. I hope you like it.)   

@Raul Simon this kind of solution I have already stated in the beginning.

If you want a new solution, here it is.

   First of all, every letter here denotes an integer. We start with a known integer N which is expressible as a sum of two squares. (It is known which integers fulfill this condition; see Davenport, "he Higher Arithmetic", London. Hutchinson, 1968, pp. 114-115.) Then, N = x squared + y squared = rs, in general. If we want to have x squared + y squared + z squared = u squared, we may write (using the previous eqs.): x squared + y squared = u squared - z squared = (u + z)(u - z) = N = rs. Then, solving the system u + z = r, u - z = s, we find: u = (r + s)/2, z = (r - s)/2. The only constraint on u, z is that they must be both odd (and, according to the theory of sums of squares, of the form 4k + 3).

     Observe that we start with a knowledge of x, y, r, s, and seek u and z. When these are found, we may write: x squared + y squared + z squared = u squared.  

Hello Raul! Hello Sujith!

In fact Raul's solution can be made even easier. Every odd natural number is the difference of two squares, because

2n+1 = (n+1)2-n2

So let x and y be two integers of different parity. Then x2+y2 is odd. Then 

x2 + y2 + n2 = (n + 1)2 

where n = (x2 + y2 - 1) / 2.

Brilliant, Mihai.  Thank you for sharing this with us. Even I can follow this algebraic reasoning,

Thank you Louis! I have looked on your paper about cubic equations. It might be of interest for you that every rational number can be written as a sum of three cubes of rational numbers (which is not true for integers!) On this task and related things, look at:

http://math.stackexchange.com/questions/494424/proving-that-any-rational-number-can-be-represented-as-the-sum-of-the-each-cube

In fact here is the 1930 article by Richmond about sums of three cubes of rational numbers:

http://journals.cambridge.org/download.php?file=%2FPEM%2FPEM2_2_02%2FS0013091500007604a.pdf&code=3012da253c5b174d50aadfcbedad50cd

This paper was beyond my level of comprehension. Have you looked at my 2006 paper? I believe one of the beauties of my paper was its relative simplicity compared to anything previously published. I'd be delighted to hear any comments anyone has concerning my paper. Thanks

Well, Richmond 's paper consists only of algebraic computations and substitutions - at least its first part - and your paper also. They are not so different, and for the moment it is not too much to understand. Better we say that understanding cames only with the computation, and is nothing else, as to become familiar with this computation. In your paper, I have better understood what is happening, when I have seen the identities (8) and (9) at page 7. Yuri Manin, a well known russian mathematician, has written a whole book of diophantine geometry about cubic forms. In the preface, he said something like: (citation by recall) "Did you know that every rational is a sum of three squares of rationals? A short proof is the following:" (and now he displays an identity with complicated terms. Then he continues:) "As you see, this proof is formally correct, but not very illuminating. For more fundamental arguments, read this book." 

So, I understood that your intuition guided you to special substitutions of variables and term manipulations, that permited you to find more solutions starting with some special solutions. Then, as I saw there, Cameron Stewart better understood your solutions and concluded that it was not so important that you have chosen x = 4m , so he has just put m instead and found out that the real reason why do the solutions work were two algebraic identities. 

The whole is a very nice story for the discovery of those nice identities. With a little bit of chance, maybe one can find a more general method to discover (verify) new pure algebraic identities by checking them on a sequence of exponentially growing values (there are already such approaches for geometric ""proof by verification"" arguments. Instead of reasoning, one computes the figure for points with coordinates which are independent enough, although they are all rational). 

But we still don't understand the things in their depth, neither by Richmond, nor by you. We understand them better by Manin, but there are still a lot of open questions and a lot of things to do. 

However, it is a very nice paper. Congrats and go on with math! (I understood that you are a doctor, so it is really nice and surprising to have maths as a hobby!)

Mihai, thank you sincerely for your thorough analysis of this question and for your very kind comments concerning my paper. I agree that there is still a lot that we don't understand about this equation. My current paper, which I will submit shortly, deals with properties of squares and cubes in relation to the solution I derived for this equation provided x is a perfect square. It does provide some interesting additional insight into this equation that hasn't been previously recognized, but at the end of the day we are no closer to being able to incorporate into our solution those solutions where x is not a perfect square. This is the largest obstacle to overcome but I doubt that it is feasible since mathematicians have tried unsuccessfully for hundreds of years  Whether the additional insight provided in my latest work is enough to make it publishable remains to be seen. Meanwhile I thank you for your encouragement and support of my work.

Well, let us suppose three natural numbers i.e. a,b and c. And choose c such that c is a multiple of a and b. Then the relation will be

a^2+b^2+(a.b)^2=x^2. Here x can be written as a*b+1. so the given equation becomes.

a^2+b^2+(a*b)^2=[(a*b)+1]^2. And a-b= +- 1.

For e.g.

2^2+3^2+6^2=7^2

3^2+4^2+12^2=13^2

and so on...