## Evens by Elap

### Puzzle explanation

The side lengths, rows, columns and one of the diagonals are square numbers in cyclic form; that is, reading digits from some point in the number to the end and then reading the remaining digits from the beginning, the result is a square number. In some cases this is just the original number; in others, two different starting points yield squares. The added lines in the solution make a 7, which is the only digit that doesn’t occur in the grid or clue values. The title is a cyclic form of SEVEN.

### Puzzle solution process

This is one possible logical path for solving the puzzle.

Since the perimeter of a square is 4 times its side length, Square 7 tells us that j2 = 4.a2, so j2 ≥ 40. For the area of Square 6 (F2.F2.j2) to be a square number, j2 must also be square. The only even 2-digit square greater than 40 is 64, so j2 = 64 and a2 = 16. The area of Square 7 is then 65536 = D4 + e2 + 64___, so D4 must be at most 65536 − 64000 − 10 = 1526, which puts a 1 in the D cell.

Similarly, Square 9 tells us that P3 = 4.t2, so P3 < 400 and is a multiple of 8. We know from the grid that P3 ends in 6 and its first digit must be even because it’s the last digit of N2 (mentioned in Square 4). Thus P3 starts with 2, being one of {216, 256, 296}, and t2 is the corresponding one of {54, 64, 74}. Using information from the grid, the maximum value for B7 is 9861898 (the 2nd digit is the last digit of s4 in Square 9; the 5th digit is the last digit of C3 in Square 1; these must be even), the maximum for U5 is 99998 and the maximum for s4 is 9888 (inner digits forced to be even by n3 in Square 4 and n4 in Square 2), so the area of Square 9 is ≤ 9971784 which makes t2 ≤ 56. Therefore t2 = 54 and P3 = 216.

The area of Square 5 is P8 + a8 + u7. From the grid, P8 is 216____, with the 5th to 8th digits all non-zero and the 5th digit even (being the last digit of f4 in Square 8), so P8 is in the range 21602112 to 21698998; a8 is 16__1_54 with the 6th digit even (the last digit of c4 in Square 3), so it’s in the range 16001054 to 16991854; u7 (which overlaps almost all of a8) is 416__1_, in the range 4160010 to 4169918. The area is then in the range 41763176 to 42860770, so the side length is in the range 6464 to 6546. As this is T2 + W4 + 16, and we know from the grid that W4 is ___4 with the 2nd and 3rd digits even, W4 must be 64_4.

For Square 6, the perimeter is M2 + u3, which is in the range 12 + 416 = 428 to 98 + 416 = 514, so the side length is in the range 108 (rounded up to be even) to 128, and the area is in the range 11664 to 16384. As this is F2.F2.64, F2 can only be 14 or 16. If F2 is 16, the perimeter is 512 and M2 is 96; that would make N4 = 6216, so the perimeter for Square 8 would be 96 + 6216 = 6312 and its area would be 2490084, which is too big for E3 + K6 + f4. Thus the only possible value is F2 = 14, which makes M2 = 32 (from the perimeter) and T2 = 58 (from the side length).

From Square 3, m3 = 4.L3, and we know m3 ends in 4, so L3 must end in 6 (not 1, since it must be even). The area is Q4 + R4 + c4, for which the maximum is 8998 + 9982 + 8892 = 27872, so the maximum side length is 166. This puts L3 in the range 106 to 166 and we can enter 1 in the L cell.

For Square 8, we have the perimeter as 32 + 2216 = 2248, so the area is 315844 = E3 + K6 + f4, which gives the first digit of K6 as 3. The side length is 562 = e2 + g3 and we now know g3 is _36, so g3 = 536 and e2 = 26. The grid now shows us E3 = 214, so K6 is in the range 315844 − 214 − 1998 = 313632 to 315844 − 214 − 1002 = 314628. That means L3 is 136 or 146, so m3 (the perimeter of Square 3) is 544 or 584 and we can write 5 in the m cell.

The side length of Square 2 is D2 + h4, where D2 = 12 and h4 is 25_4 with the missing digit even, so it’s in the range 12 + 2504 = 2516 to 12 + 2584 = 2596, and the area is in the range 6330256 to 6739216. This is S4.h4 + S5, where S4 is _558 and S5 is _5584. If the first digit of S4 and S5 is 1, the maximum area would be 1558.2584 + 15584 = 4041456, which is too small. If the first digit is 3, the minimum area would be 3558.2504 + 35584 = 8944816, which is too big. So S4 = 2558, S5 = 25584. Of the five possibilities for h4, only h4 = 2544 gives a square number for the area, 6533136, giving a side length of 2556 which agrees with 12 + 2544.

The grid now tells us m3 = 544, so L3 = 136 (side of Square 3), which makes K6 = 3136__. From the area of Square 8, the minimum value of f4 is now 315844 − 214 − 313698 = 1932 and we know it’s 1_3_, so its 2nd digit is 9. That means the minimum value of K6 is now 315844 − 214 − 1938 = 313692, so its 5th digit is 9.

For Square 7, we have the area 65536 = D4 + e2 + j5 = 1214 + 26 + j5, so j5 = 64296, which includes p4 = 4296. The perimeter of Square 2 is 4.2556 = 10224, which is n4 + 4296, so n4 = 5928.

We know the area for Square 9 is 8503056 = B7 + U5 + s4. The grid tells us B7 is __61214, U5 is _2558 and s4 is 928_, so s4 = 9284 (so the last digits add up), which makes B7 = _461214. B7 must be in the range 8503056 − 92558 − 9284 = 8401214 to 8503056 − 12558 − 9284 = 8481214, so B7 = 8461214 and U5 = 32558.

The area of Square 4 is 1536.k3.r2, where k3 starts with an even digit (the last digit of f4 in Square 8) and r2 is the last two digits of k3. The maximum value for the area is 1536.898.98, so the side is ≤ 11626 and the perimeter (b2.1536) is ≤ 46504, so b2 ≤ 30. We know from the grid that b2 is _4 with the first digit even (the last digit of A6 in Square 1), so b2 = 24, the perimeter is 36864 and the area is 84934656. Now k3.r2 = 84934656 / 1536 = 55296. If we express k3 as 200x + 2y, where x is 1 to 4 and y is 1 to 49, then r2 = 2y and (200x + 2y)2y = 55296, which can be reduced to (100x + y)y = 13824. We then want a pair of factors that multiply to make 13824 (= 29.33) and whose difference is a small multiple of 100. Since the factors end in the same digit, it must be 2 or 8. The only pairs with a difference in the range 100 to 400 are 192.72, 288.48 and 432.32, so x = 4 and y = 32, which makes r2 = 64 and k3 = 864. The grid now tells us f4 = 1938, so K6 = 313692 (= 315844 − 214 − 1938), using the area of Square 8.

For Square 3, we know the area is 18496 = Q4 + R4 + c4, where Q4 is 8_32, R4 is _322 (overlapping with Q4) and c4 is 4__2 (with the 2nd digit even). Looking at the last two digits, c4 must end in 96 − 32 − 22 = 42. R4 is in the range 18496 − 8932 − 4842 = 4722 to 18496 − 8132 − 4042 = 6322, but we know it matches _322, so R4 is either 5322 or 6322. Then c4 is either 18496 − 8632 − 6322 = 3542 or 18496 − 8532 − 5322 = 4642, but we know it matches 4_42, so c4 = 4642, R4 = 5322 and Q4 = 8532.

For Square 5, the side is 58 + 6424 + 16 = 6498, the perimeter is 25928 + 64 = 25992 and the area is 42224004 = P8 + a8 + u7, for which we have 216_8532 + 16__1854 + 416__18, with the missing digits in the latter two numbers identical. To complete the sum, P8 = 21698532, a8 = 16361854 and u7 = 4163618.

For Square 4, the side is 9216 = 22.q3 + 592, so q3 = 392.

For Square 1, the area is in the range 816002 + 532216 + 1536 = 1349754 to 816992 + 532216 + 1536 = 1350744. The only square number in that range is 1350244, so A6 = 816492 and the side is 1162. The perimeter is 4648 = D2.H2 + J4, where D2 = 12, H2 is _6 and J4 is ___6 (the last two digits being H2). If we express J4 as 1000x + 100y + 10z + 6, where x, y and z are digits in the range 1 to 9 (all non-zero because they start clued numbers), then H2 = 10z + 6. The perimeter gives 12(10z + 6) + (1000x + 100y + 10z + 6) = 4648, which simplifies to 100x + 10y + 13z = 457. The first two terms end in 0, and the only digit that gives a product ending in 7 when multiplied by 13 is 9, so z = 9. This gives 100x + 10y + 13.9 = 457, so 10x + y = 34, which means x = 3 and y = 4. Thus J4 = 3496, G3 = 496 and H2 = 96. The side gives 1162 = 612 + 496 + V2, so V2 = 54.

The clues are now all solved and the grid is complete except for three cells each in the top-left and bottom-right corners. Combining the themes of squares in the clues and cyclic entry in the grid, the thematic property is that numbers are squares in some cyclic permutation. For example, the side of Square 1 is 1162, which with its last digit moved to the front gives 2116 = 462. Three of the sides, 9216, 256 and 2916, are squares already, so their “permutation” is the one that starts with the first digit. Square 8’s side of 562 can be cycled to give 625 = 252 and 256 = 162. The eight complete rows and columns of the grid also have this property. For example, if we read the third row starting from the last cell (marked J) then cycling to the start of the row, we get 34963569 = 59132.

Any square must end with one of {0, 1, 4, 5, 6, 9}. If it ends with 0 it has a factor of 10 and, being square, it must therefore have a factor of 100, so it must end with 00. If it ends with 5 it has a factor of 5 and therefore has a factor of 25, so the last two digits must be 25 or 75. If it ends with 4 or 6 it has a factor of 2 and therefore has a factor of 4, so the last two digits must be a multiple of 4.

In column 7, the potential squares are 9193864_, 193864_9, 93864_91, 3864_919 and _9193864. Trying 0 for the missing digit, √91938640 is a bit more than 9588, and 95892 = 91948921, which is greater than 91938649, so there’s no matching square; √19386409 = 4403, so we’ve found a square (the next square ending in 9 is 44072 = 19421649, so there are no other possibilities for the missing digit); √93864091 > 9688, and 96892 = 93876721 > 93864991 (no match); √38640919 > 6216, and 62172 = 38651089 > 38649919 (no match). That leaves _9193864, for which none of the possibilities are square. Thus column 7 is 91938640.

Row 8 is now 6424920_, so the potential squares are 6424920_, 424920_6, 24920_64, 920_6424 and 20_64249. In turn, √64249200 > 8015, and 80162 = 64256256 > 64249209 (no match); √42492006 > 6518, and 65192 = 42497361 > 42492096 (no match); √24920064 = 4992 (the next square ending in 4 is 49982 = 24980004, so there are no more possibilities for the missing digit); √92006424 > 9591, and the next squares ending in 4 are 95922 = 92006464 and 95982 = 92121604 > 92096424 (no match). That leaves 20_64249, for which none of the possibilities are squares. Thus row 8 is 64249200.

Column 8 is now 243653_0, so the potential squares are 243653_0, 3653_024, 53_02436 and 0243653_. In turn, √24365300 > 4936, and 49372 = 24373969 > 24365390 (no match); √36530024 > 6044, and 60452 = 36542025 > 36539024 (no match); for 53_02436, only 53202436 = 72942 is a square; √02436530 > 1560, and 15612 = 2436721 > 02436539 (no match). Thus column 8 is 24365320. Row 7 is now complete as 54952942, so the potential squares are 49529425, 52942549 and 42549529. Only the last of these is a square, with √42549529 = 6523.

In row 2, the potential squares are 8461214_, 61214_84, 214_8461 and 4_846121. In turn, √84612140 > 9198, and 91992 = 84621601 > 84612149 (no match); √61214084 > 7823, and the next square ending with 4 is 78282 = 61277584 > 61214984 (no match); √21408461 > 4626, and the next squares ending with 1 are 46292 = 21427641, 46312 = 21446161, 46392 = 21520321 > 21498461 (no match). That leaves 4_846121, for which only 48846121 = 69892 is square, so row 2 is 88461214.

In column 2, the potential squares are _8922544, 8922544_, 22544_89 and 44_89225. In turn, none of the possibilities for _8922544 are square; √89225440 > 9445, and 94462 = 89226916 > 89225449 (no match); √22544089 > 4748, and 47492 = 22553001 > 22544989 (no match); of the possibilities for 44_89225, only 44689225 = 66852 is square. Thus column 2 is 68922544.

In column 1, the potential squares are now _8493256, 8493256_, 93256_84, 3256_849 and 6_849325. In turn, none of the possibilities for _8493256 are square; √84932560 > 9215, and 92162 = 84934656 > 84932569 (no match); √93256084 > 9656, and 96572 = 93257649 > 93256984 (no match); √32560849 > 5706, and 57072 = 32569849 (match); none of the possibilities for 6_849325 is square. Thus column 1 is 98493256.

Row 1 is now complete and the potential squares are 68164929, 81649296, 64929681, 49296816, 92968164 and 29681649, of which both only 81649296 = 90362 and 92968164 = 96422 are square. Thus row 1 is the one that is doubly thematic. The diagonal from bottom-left to top-right is 64513612, which is a cycled form of 12645136 = 35562. Lines drawn through this diagonal and row 1 make the shape of a 7, which is the only digit that doesn’t occur anywhere in the grid or the values used in the clues.