At the start, **A1=4**.

Since each even number other than 2 needs to be adjacent to its half, we know that 4 borders 2. 2 cannot be in the B column as it is a prime, so **A2=2**.

8 must be adjacent to 4 and by the clue in A2, 8 cannot be in row 1 as it is a cube number. Thus, **B2=8.**

16 must also be adjacent to 4 as it is its root, so **B1=16.**

A5 must be a square number which narrows it down to 1, 9, or 25. However, 9 and 25 both have roots that are prime and odd, thus cannot be in column B or A4 respectively. Thus, **A5=1.**

Since the numbers in column E are in descending order and the two remaining square numbers are 9 and 25, **E1=25** and **E5=9**.

Since 5 must be adjacent to 25, Row A is in ascending order, and column E is in descending order, **D2=5.**

Now consider the number 7. It cannot be in Row 1 or Column E due to the ascending and descending orders. Tt cannot be in Column B as it is a prime. It cannot be in A3, C2, or C3 as that would be adjacent to the number 8. It cannot be in A4 as it is odd. The 3 must be in either D4 or D5 as it must be adjacent to 9 and not in Column E (due to the descending order). The 6 must border the 3 and it cannot be in C3 or D3 as that would border 5. It also cannot be in E4 because of the descending order. Thus, the 6 must be in the square consisting of C4, C5, D4, and D5. 7 cannot be in that square as it would border 6. Thus, by process of elimination, **D3=7. **

C1 must be 19, 20, or 21 as, from the set of numbers between 16 and 25, 17 is consecutive to 16, 18 must border 9, and 22-24 would result in making it impossible to fill in D1. D1 must be 21 or 23 as 20 would border 19, 24 would border 25, and 22 would force C2 to be 11, making it impossible to place the 10 adjacent to the 5.

Now consider the number 6. It had been previously narrowed down to four squares but two of them, C4 and D4, can further be eliminated because they border the 7. That leaves C5 and D5. Assume that D5 is 6. That would force D4 to be 3 which would force E4 to be 18 as to be adjacent to the 9. This forces E3 to be 20 as 19 is consecutive to 18, 21 would result in E2 being 23 resulting in nothing possible to place in D1, and both 23 and 24 would make it impossible to fill E2. However, if that were the case, there would be nowhere to place the 10 as it must border both 5 and 20 and it would not fit in E2 due to the descending order. Thus, the 6 is not in D5. By process of elimination, **C5=6.**

Because 3 is odd, it cannot be in D4. By process of elimination, **D5=3.**

The same logic that proved that 6 was not in D5 proves that 18 is not in E4. 18 must border 9 so by process of elimination, **D4=18.**

Now consider E4. It must be 13, 14, or 15 as 16 and 18 are already placed elsewhere, 17 and 19 are consecutive to 18, 11 would need to be next to 22 which would require 22 to be in E3 which would not allow anything to be placed in E2, 12 needs to be adjacent to 6, and 20 or greater makes it impossible to fill E2. Thus, the lowest number that can be in E3 is 15, as it must be at least 2 greater than E4. E4 cannot be 16 or 18 as they are placed elsewhere and cannot be 17 or 19 as they are consecutive to 18. E4 cannot be 20 as there would be nowhere to place the 10 that borders both 5 and 20. It cannot be 21 as that would force E2 to be 23, leaving nothing for D1. 22, 23, and 24 all make it impossible to fill E2. Thus, by process of elimination, **E3=15.**

Since E3 is 15, E4 cannot be 15 or 14, so **E4=13.**

Now consider the number 11. It cannot be in Row 1 or column E due to the ascending and descending order. The number 10 must be in either C2 or C3 to border 5 and not be in Row 1 or column E, so 11 cannot be in C2 or C3 as it is consecutive to 10. Similarly, the number 12 must be in B4, B5, or C5 to border the number 6 so the 11 cannot be in C4 as it is consecutive to 12. It also cannot be in column B as it is prime and cannot be in A4 as it is odd. Thus, by process of elimination,

**A3=11.**

Now consider the number 17. It cannot be in A4 or in column B as it is odd and prime respectively. It cannot be in C1 or C2 as it is consecutive to 16 and it cannot be in C3 or C4 as it is consecutive to 18. It cannot be in D1 because that would not allow anything to fill C1. Thus, by process of elimination, **E2=17.**

Now consider the number 19. It cannot be in A4 or column B because it is odd and prime respectively. It cannot be in C3 or C4 as it is consecutive to 18 and cannot be in D1 as it would make it impossible to fill C1. Thus, 19 is in C1 or C2.

Now consider the number 21. It cannot be in A4, B3, or B4 as it would be forced to border 22 which must be in one of those to be adjacent to 11. B5 cannot be 21 as it would force B3 to be 22, which would force B4 to be 24 as that would be the only non-prime number left that would fit, which would leave no even numbers that could go in A4 as 10, 12, and 14 must be adjacent to numbers elsewhere. Thus, 21 is not in B5. 21 is not in C3 or C4 as they both would prevent column B from being filled up as they eliminate the number 21 (because it would be used) and the numbers 20 and 22 from being in the column (neither can be in B5 because 22 must border 11 and 20 must border 10 which must border 5). Thus, 21 must be in either C1, C2, or D1.

Now consider the number 23. It cannot be in A4 or Column B because it is odd and prime respectively. It cannot be in C4 as it would make it impossible to fill Column B by eliminating the numbers 22 and 24 from being in it. It cannot be in C3 because it is odd and cannot be in C1 because it would make it impossible to fill D1. Thus, 23 is in either C2 or D1.

Taking the last three paragraphs together, 19, 21, and 23 must fill C1, C2, and C3. Thus, 10, which must border 5, has only one location left. **C3=10.**

14 must border 7 and there is only one location left. **C4=14.**

12 must border 6. It cannot be in B4 as it is consecutive to 11. Thus,

**B5=12.**

24 must border 12. 24 cannot be in A4 as it is a multiple of 3. Thus,

**B4=24.**

20 must border 10. It cannot be in C2 as that is in the 19, 21, 23 group of three. Thus, **B3=20.**

22 must border 11 and there is only one space left for that. **A4=22.**

Since C2 borders 20, it cannot be 19 or 21. Thus, **C2=23.**

Since Row 1 is ascending, **C1=19** and **D1=21.**

1-25 Logic Puzzle by Freesquirrel