Problem of the Month(December 2015)

What are the smallest two squares that together contain the squares of sides 1 through n? More precisely, how small can the total area be of at most two squares containing non-overlapping copies of these squares be? Does one of the two optimal squares always contain no more than 1 square?

What are the answers if we consider squares with areas 1 through n instead of sides 1 through n? 2:   22 + 12 = 5 3:   32 + 32 = 18 4:   52 + 42 = 41 5:   72 + 52 = 74 6:   92 + 62 = 117 7:   132 = 169 8:   152 = 225 9:   152 + 92 = 306 10:   182 + 92 = 405 11:   212 + 92 = 522 12:   242 + 102 = 676 (MM) 13:   272 + 122 = 873 (MM) 14:   302 + 132 = 1069 (MM) 15:   332 + 142 = 1285 (MM) 16:   392 = 1521 (MM) 17:   392 + 172 = 1810 (GS) 18:   442 + 142 = 2132 (GS) 19:   472 + 172 = 2498 (MM) 20:   512 + 172 = 2890 (MM) 21:   562 + 142 = 3332 (MM) 22:   602 + 152 = 3825 (MM) 23:   642 + 162 = 4352 (MM) 24:   662 + 242 = 4932 (MM) 25:   722 + 202 = 5584 (MM) 2:   √22 + √12= 3 3:   (√2+√1)2 + √32= 8.828+ 4:   (√3+√2)2 + √42= 13.898+ 5:   (√5+√4)2= 17.944+ 6:   (√5+√4)2 + √62= 23.944+ 7:   (√4+√3+√2)2 + √52= 31.484+ 8:   (√5+√4+√3)2 + √62= 41.618+ 9:   (√6+√5+√4)2 + √72= 51.696+ (MM) 10:   61.2+(DC) 11:   (√8+√7+√6)2 + √92= 71.784+ 12:   (√10+√9+√8)2= 80.832+ (DC) 13:   (√13+√8+√7)2 + √112= 93.441+ (MM)