|
Note that if two or more of the squares have the same value, then two or more of the sides must also have the same value, contrary to the stipulation of the problem.
One possible starting point is to set p,q,r,s in arithmetic progression. The minimal such progression is:
a+b+c = (x)^2
a+b+d = (x-1)^2
a+c+d = (x-2)^2
b+c+d = (x-3)^2
giving, per WolframAlpha:
a = 3n^2+16n+14, b = 3n^2+10n+9, c = 3n^2+4n+2, d = 3n^2-2n-7.
and also
a = 3n^2+8n-2, b = 3n^2+2n+1, c = 3n^2-4 n+2, d = 3n^2-10 n+1.
There are infinitely many different such dice, but checking small values gives the smallest possible set as {1,22,41,58}.
If the summed squares of such a die differ by 1, then all four sides cannot sum to a square. Adding a,b,c,d in the first case gives:
3n^2+16n+14+3n^2+10n+9+3n^2+4n+2+3n^2-2n-7 = 12n^2+28n+18=a^2;
and in the second:
3n^2+8n-2+3n^2+2n+1+3n^2-4n+2+3n^2-10n+1 = 12n^2-4n+2=a^2
neither of which has Diophantine solutions in {n,a}.
However, solution sets are also available for other arithmetic differences between p, q, r, and s, such as (x-1)^2, (x-3)^2, (x-5)^2, (x-7)^2, and (x^2), (x-2)^2,(x-4)^2, (x-6)^2, etc.
Nor is it essential that the differences be arithmetic. In particular, the set (x)^2, (x-1)^2, (x-2)^2, (x-6)^2 has as a solution: 16(12n^2+14n+1), 8(24n^2+4n+1), 192n^2-16n+1, 8(24n^2-8n-1). Adding these gives 768n^2+176n+17 = a^2, which does produce a (recurrent) solution, the smallest value of a being 31; the next value is considerably larger, with x = 42022, a = 1164551. So there are infinite ways in which to satisfy Alice.
|
