A344332 Side s of squares of type 2 that can be tiled with squares of two different sizes so that the number of large or small squares is the same.

15, 30, 45, 60, 65, 75, 90, 105, 120, 130, 135, 150, 165, 175, 180, 195, 210, 225, 240, 255, 260, 270, 285, 300, 315, 325, 330, 345, 350, 360, 369, 375, 390, 405, 420, 435, 450, 455, 465, 480, 495, 510, 520, 525, 540, 555, 570, 585, 600, 615, 630, 645, 650, 660, 671, 675, 690, 700, 705, 715, 720, 735, 738, 750, 765, 780, 795, 810, 825, 840, 845, 855, 870, 875, 885, 900

Offset: 1

Author

Bernard Schott, May 20 2021

Keywords

• nonn

Comments

This sequence is relative to the generalization of the 4th problem proposed for the pupils in grade 6 during the 19th Mathematical Festival at Moscow in 2008 (see A344330).

There are two types of solutions, the second one is proposed here, while type 1 is described in A344331.

If m is a term and k > 1, k * m is another term.

Every term (primitive or not primitive) is the side of an elementary square of type 2 (see A346263).

Some notations: s = side of the tiled square, a = side of small squares, b = side of large squares, and z = number of small squares = number of large squares.

-> Primitive squares

Side s of primitive squares of type 2 must satisfy the Diophantine equation s^2 = z * (a^2+b^2) with the conditions a^2+b^2 = c^2 and gcd(a, b, c) = 1.

In this case, q = a/(c-b) must be odd, and side s = q*c = a*c/(c-b) = (a+b)*c/a with a number of squares z = q^2 = (a/(c-b))^2 = ((b+c)/a)^2.

Indeed, these conditions give exactly the following solutions for n >= 2: s = n^4-(n-1)^4 (A005917), a = 2*n-1 (A005408), b = 2*n*(n-1) (A046092), c = 2*n*(n-1)+1 (A001844), z = (2*n-1)^2 (A016754); this results come from the identity:

(n^4 - (n-1)^4)^2 = (2*n-1)^2 * ((2*n-1)^2 + (2*n*(n-1))^2).

For n >= 2, every primitive square is composed by a square S1 of z = (2*n-1)^2 large squares with side b = 2*n*(n-1), then an edge on two sides of this square S1 of z = (2*n-1)^2 small squares with side a = 2*n-1.

See example with design of square of side s = 15 with a = 3, b = 4, c = 5, q = 3, z = 9, obtained with n= 2.

-> Non-primitive squares

If s is the side of a primitive square of type 2, then every k * s, k > 1 is a non-primitive term that gives two distinct tilings of type 2.

The square ks X ks can be tiled with z = q^2 = (2n-1)^2 = (a/(c-b))^2 = ((b+c)/a)^2 squares of side ka and of side kb, but also,

The square ks X ks can be tiled with z = k^2*q^2 = ((2n-1)*k)^2 = (k*a/(c-b))^2 = (k*(b+c)/a)^2 squares of side a and of side b (see example).

Examples

			Primitive square with s = 15:
  a = 3, b = 4, c = 5, s = 15, z = 9; s = 15 is the side of primitive square, with  z = 9 squares of size 3 x 3 and 9 squares of size 4 x 4
Non-primitive square k*s = 2*15 = 30:
  a = 3, b = 4, c = 5, s = 30, z = 36, this square is obtained with 4 copies of the primitive square as below.
  a = 6, b = 8, c = 10, s = 30, z = 9, this square and its tiling are exactly as the primitive square with scale 2.
               b = 4 (or = 8)     a = 3 (or = 6)
            ________ ________ ________ ______ ______________________________
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           |_____ __|___ ____|_ ______|______|                              |
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           |_____|______|______|______|______|______________________________|
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                      s = 15               s = 30
		

References

• Ivan Yashchenko, Invitation to a Mathematical Festival, pp. 10 and 102, MSRI, Mathematical Circles Library, 2013.

Links

• Index to sequences related to Olympiads.

Crossrefs

Cf. A001844, A005408, A005917, A046092, A046754, A344330, A344331.

Programs

PARI

pts(lim) = my(v=List(), m2, s2, h2, h); for(middle=4, lim-1, m2=middle^2; for(small=1, middle, s2=small^2; if(issquare(h2=m2+s2, &h), if(h>lim, break); listput(v, [small, middle, h])))); vecsort(Vec(v)); \\ A009000
isdp4(s) = my(k=1, x); while(((x=k^4 - (k-1)^4) <= s), if (x == s, return (1)); k++); return(0);
isokp2(s) = {if (!isdp4(s), return(0)); if (s % 2, my(vp = pts(s)); for (i=1, #vp, my(vpi = vp[i], a = vpi[1], b = vpi[2], c = vpi[3]); if (a*c/(c-b) == s, return(1)); ); ); }
isok2(s) = {if (isokp2(s), return (1)); fordiv(s, d, if ((d>1) || (dMichel Marcus, Jun 04 2021