20230318, 21:22  #1 
May 2003
7·37 Posts 
squarewise concatanation for the win
1225172265625 is a lovely number
1 is a square 225 is a square put them together, and 1225 is a square 72265625 is a square put it together with what came before, and you get 122572265625 which is a square I don't know where I'm going with this, but I'm sure you guys can have fun. (oh, 4 and 9 have even shorter paths, but ones I've not probed as deeply) 
20230319, 13:12  #2  
Feb 2017
Nowhere
13·487 Posts 
Quote:
Quote:
Code:
? print(factor(72265625)) [5, 9; 37, 1] Quote:
Code:
? print(factor(122572265625)) [3, 2; 5, 9; 19, 1; 367, 1] 

20230319, 13:54  #3  
May 2003
7×37 Posts 
Quote:
? factor(172265625) [3 2] [5 8] [7 2] ? factor(1225172265625) [ 5 8] [ 7 2] [11 2] [23 2] 

20230319, 14:21  #4  
May 2003
100000011_{2} Posts 
Quote:
4 > 49 > 49729 > 4972922562328439750054749703581033600295913998934338451363146305084228515625 9 > 950625 > 95062561096409435940120435784167300872787977524634772310552222052137949503958225250244140625 

20230322, 14:11  #5 
Feb 2017
Nowhere
18BB_{16} Posts 
16, 1681, 1681236390625
Code:
4, 49, 49729, 497299167114067888529849495363612005139078505485784844495356082916259765625 Code:
9, 93025, 93025698201261719352105289560261241444780223428635104730773756605244766299827431245265431380975229558316641487181186676025390625 Last fiddled with by Dr Sardonicus on 20230322 at 17:18 Reason: Additional examples 
20230322, 17:34  #6  
May 2003
7·37 Posts 
Quote:
Can I enquire your algorithm? Mine's an idiotic brute force: To extend N by up to D_max digits: For each length D <= D_max Try to extend N by D digits To extend N by D digits: For each pair L,H : L*H=N Look at the set of values L*2^x*5^y and H*2^(Dx)*5^(Dy), finding ones that are close to each other If their difference is small enough, a solution can be found. The only smarts I've added are traversing the 2D grid of x,y values by following a diagonal so that for each y you never look at more than two y values, as you zigzag along. There's hopefully a proper mathematical solution that's better than this almost completely dumb brute force. 

20230322, 20:09  #7  
Feb 2017
Nowhere
13·487 Posts 
Quote:
So x  y and x + y are complementary divisors of N*10^D. The inequalities force y to be rather small, so x + y and x  y will be just either side of the square root of N*10^D. The divisors x  y and x + y also have to be of the same parity. I started just past the middle divisor and looked at successively larger divisors. You don't have to look far. As soon as two complementary divisors differ by more than 2*10^(D/2) you're done. 

20230323, 13:30  #8 
Feb 2017
Nowhere
13×487 Posts 
I do not know whether every square can be concatenated on the right with a square to form another square. If it is possible, of course, the operation could (in theory) be continued indefinitely starting with any given square. It was also not clear to me whether a given square has infinitely many "square right concatenands" irrespective of whether they are themselves concatenations of squares.
I found the "least right square concatenands" for n^2, n = 1 to 100. For n = 1 to 32 we have (n, n^2, concatenand) Code:
1 1 225 2 4 9 3 9 3025 4 16 9 5 25 889914313729282379150390625 6 36 1 7 49 729 8 64 144101272782851806640625 9 81 225 10 100 126609965386962890625 11 121 84462890625 12 144 4 13 169 2601 14 196 11026550390625 15 225 625 16 256 14402025 17 289 1625853603515625 18 324 9 19 361 56169 20 400 225031640625 21 441 73530625 22 484 27228828515625 23 529 2976043447265625 24 576 9604 25 625 3516119384765625 26 676 50625 27 729 102515625 28 784 11025 29 841 118265625 30 900 1265625 31 961 38025 32 1024 144 Code:
93992319734581661956387986883358273574988380195353931821736541786064242089305750894379955229229750557351508177816867828369140625 = 9694963627295445503856592180983186990852118469774723052978515625^2 1000093992319734581661956387986883358273574988380195353931821736541786064242089305750894379955229229750557351508177816867828369140625 = 1000046995055599665423155971380983186990852118469774723052978515625^2 Consider the square 1. Concatenating on the left by the square 36 gives 361 = 19^2. But it is clear that there are infinitely many "left square concatenands" for 1. They are the solutions to the quadratic Diophantine equation 10*b^2 + 1 = a^2, which can be characterized by Mod(a + b*x, x^2  10) = Mod(19 + 6*x, x^2  10)^i, i = positive integer. For i = 2 we get 10*228^2 + 1 = 721^2. Multiplying through by 4 gives all "left concatenands" for 4. For the square 9, multiplying the solutions for the square 1 gives some, but not all, "left concatenands" for 9. There are two other families of solutions, beginning with 49 and 169, characterized by Mod(a + b*x, x^2  10) = Mod(7 + 2*x, x^2  10)*Mod(19 + 6*x, x^2  10)^i, i = nonnegative integer and Mod(a + b*x, x^2  10) = Mod(13 + 4*x, x^2  10)*Mod(19 + 6*x, x^2  10)^i, i = nonnegative integer. The next members of these families are 10*80^2 + 9 = 253^2 and 10*154^2 + 9 = 487^2 . Last fiddled with by Dr Sardonicus on 20230323 at 14:23 