Timeline for Consecutive Integer Squared Square
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 22, 2014 at 16:25 | comment | added | individ | I decided on a different way to write the solution. These formulas do not give all solutions. For the equation: $$n^2+(n+1)^2+(n+2)^2+...(m-2)^2+(m-1)^2+m^2=q^2$$ If we choose this number " $t$ " so it was odd and it was not a multiple of 3. Then for and another solution: $$k=\frac{t^2-1}{6}$$ Lay on multipliers and find $z,a$. $$k(3k+1)=(z-a)(z+a)$$ Then $$q=zt$$ $$m=a+3k$$ $$n=a-3k$$ It is necessary that $ " n " $ was greater than zero. Otherwise there will be confusion. | |
| Oct 21, 2014 at 13:17 | answer | added | Stuart Anderson | timeline score: 3 | |
| Feb 5, 2014 at 0:13 | comment | added | Gerry Myerson | 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). | |
| Feb 4, 2014 at 22:56 | history | edited | user9072 |
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| Jun 25, 2013 at 3:02 | review | First posts | |||
| Jun 26, 2013 at 6:43 | |||||
| Jun 14, 2013 at 17:33 | answer | added | Nils Rosehr | timeline score: 6 | |
| Jun 11, 2013 at 13:55 | comment | added | j.c. | 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 | |
| Jun 11, 2013 at 4:58 | comment | added | Noam D. Elkies | 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. | |
| Jun 11, 2013 at 4:10 | comment | added | Matt Watson | Is there any generalization with k,...n? | |
| Jun 11, 2013 at 4:00 | answer | added | Gerry Myerson | timeline score: 11 | |
| Jun 11, 2013 at 3:37 | comment | added | Noam D. Elkies | 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. | |
| Jun 11, 2013 at 3:14 | history | asked | Matt Watson | CC BY-SA 3.0 |