Is it possible to construct a squared square out of consecutive integer squares? Be it 1,2,3,...n or k,k+1,k+2,...n.

According to James R Bitner and Edward M Reingold, Backtrack programming techniques, Communications of the ACM 18 (November 1975) 651656, it [EDIT: meaning, the $1,2,\dots,n$ case] can't be done. "As a final example, consider the problem of tiling a $70\times70$ square with square tiles of sizes 1 through 24.... The identity $$1^2+2^2+\cdots+24^2=4900=70^2$$ suggests ... it may be possible, but a backtrack with macros program demonstrated that it is not." 


On Squaring.Net there is a comprehensive list of all simple perfect squared squares up to order 30, that is squares that can be covered by up to 30 squares. A search reveals that among those the following comes closest to the question: the square with sidelength 332 can be covered in five distinct ways with 30 squares, two of which use all the squares of sidelengths 2 to 12, i.e. 11 consecutive integer squares. There are only 5 more squares (of orders 28, 30 and 27) using more than 7 squares of consecutive integer sizes. Also interesting is one of the order 22 cases, which uses only 7 series (which is minimal) of consecutive integer squares to cover a square of side length 110; this also happens to be the smallest square which can be covered at all. 


If a solution was possible to tile a 70x70 square with squares $$1^2+2^2+\cdots+24^2$$ then a perfect squared square of order 24 would exist. Order 24 has been completely enumerated. The compound perfect squared squares (CPSSs  compound; subrectangles allowed) of order 24 were generated by Duijvestijn, Federico and Leeuw in 1979 and 1982. Only 1 CPSS of order 24 was found, T.H. Willcocks 175x175, discovered in 1948. The 26 simple perfect squared squares (SPSSs  simple; no subrectangles allowed) of order 24 were enumerated by Duijvestijn in 1991, the smallest SPSS of that order being 120x120. These results have been confirmed a number of times by other researchers using different approaches and software. The sum of squares formula gives;$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$ $$For\ n = 32,\ \sum_{i=0}^n i^2 = 11440, \sqrt{11440}= 106.95794..$$ $$For\ n = 33,\ \sum_{i=0}^n i^2 = 12529, \sqrt{12529}= 111.93301..$$ If perfect squared squares existed with a side less than 110 in length then they would need to be of order 32 or less (if the 33 squares from 1 to 33^2 will not fit in a square of side 110, neither will a larger set of 33 squares, nor will any order higher than 33). All orders of perfect squares squares up to and including order 32, both simple and compound, have been enumerated, (order 33 has also been done this year by Jim Williams) see www.squaring.net. The smallest perfect squared squares found up to order 33 were the 3 SPSSs with sides of 110. There are 2 in order 22, and 1 in order 23. So these are the smallest possible perfect squared squares . Note that the 112 side SPSS found by Duijvestijn in 1978 is the lowest order (21) perfect squared square, but not the smallest. Gambini achieved the same result, that 110 is the smallest perfect squared square in his 1999 thesis, using a packing program. I do not know of any consecutive square tiling, unless more than 1 of each square is allowed. http://www.mathpages.com/home/kmath147.htm Sum of Consecutive Nth Powers Equals an Nth Power, has a generalisation to cubes. 


For the equation: $$n^2+(n+1)^2+(n+2)^2+...(m2)^2+(m1)^2+m^2=z^2$$ If you choose any odd number " $t$ " And lay multipliers. $$(t^21)(t^2+1)=12(qtk)(qt+k)$$ This formula will find $  "q,k"$ Then a fairly simple solution, you can write: $$m=k+\frac{t^21}{2}$$ $$n=k\frac{t^2+1}{2}+1$$ $$z=qt^2$$ It is necessary that $ " n " $ was greater than zero. Otherwise there will be confusion. 

