Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
4  
For $1,2,3,\ldots,n$ it's known that $n$ would have to be $24$ just because that's the only nontrivial solution of the Diophantine equation $1^2+2^2+3^2+\ldots+n^2 = y^2$. I've seen the claim that a $70 \times 70$ square cannot be tiled with squares of sides $1,2,3,\ldots,24$, but I don't know if that's actually been proved. I've also seen a picture where all these squares except the $7 \times 7$ are packed into the $70 \times 70$, but I don't remember the reference, nor whether that's supposed to be as much of the $70 \times 70$ square that can be covered by a subset of those $24$ squares. –  Noam D. Elkies Jun 11 '13 at 3:37
    
Is there any generalization with k,...n? –  Matt Watson Jun 11 '13 at 4:10
2  
There are lots of Diophantine solutions: telling gp "f(n) = n*(n+1)*(2*n+1)/6; for(i=1,100,for(j=0,i-2,if(issquare(f(i)-f(j),&s),print([j+1,i,s]))))" yields the following list (sorted by $s$): $$ $$ [3,4,5] [20,21,29] [1,24,70] [18,28,77] [7,29,92] [9,32,106] [17,39,138] [7,39,143] [38,48,143] [20,43,158] [25,48,182] [25,50,195] [7,56,245] [27,59,253] [44,67,274] [25,73,357] [28,77,385] [22,80,413] [76,99,430] [60,92,440] [44,93,495] [38,96,531] $$ $$ Some are impossible because they'd yield squared squares of order $<21$, but there are still quite a few candidates. –  Noam D. Elkies Jun 11 '13 at 4:58
    
You might like this article on tiling the plane with squares of whole number sides, one of each size math.smith.edu/~jhenle/stp/stp.pdf –  j.c. Jun 11 '13 at 13:55
1  
The numbers $m^2$, $m\le2828$, that can be expressed as a sum of two or more consecutive squares, are tabulated at oeis.org/A097812 (tiling is not discussed). –  Gerry Myerson Feb 5 at 0:13

3 Answers 3

According to James R Bitner and Edward M Reingold, Backtrack programming techniques, Communications of the ACM 18 (November 1975) 651-656, 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."

share|improve this answer
1  
Martin Gardner reports on this on page 148 of Mathematical Carnival, and adds that his readers sent 27 examples packing all but the $7\times7$ into the $70\times70$. He gives an example on page 149. He reports that it remains a conjecture that missing 49 square units is the best you can do. But perhaps there have been improvements in the intervening years. –  Gerry Myerson Jun 11 '13 at 4:14
4  
Found it. R. Korf, Optimal rectangle packing: New results, in Proc 14th Internat Conf on automated planning and scheduling (2004) 142-149, also Korf, Moffitt, Pollack, Optimal rectangle packing, Ann Oper Res 179 (2010) 261-295, it's proved that missing 49 square units is the best possible. –  Gerry Myerson Jun 11 '13 at 4:22
    
Thanks! Yes, the Gardner report must be what I saw. I didn't know of the Korf computation. –  Noam D. Elkies Jun 11 '13 at 4:53

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 side-length 332 can be covered in five distinct ways with 30 squares, two of which use all the squares of side-lengths 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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.