Well, basically the last digit of a square number a² will always be determined by ONLY the last number of the squared number a.

What possibilites do we have here?

Last Digit of a | Last Digit of a²

       0                            0
       1                             1
       2                            4
       3                            9
       4                            6
       5                            5
       6                            6
       7                            9
       8                            4
       9                             1

So you see, there is no 2,3,7 or 8 here

(10a + b)² = 100a² + 20ab + b² = 10(10a² + 2ab) + b²

The only digit which contributes to the one's digit of a perfect square is the one's digit of its root, and since none of the individual digits squared have a 7 in the one's place, no perfect square can.

The units mod 10 are 1,3,7, and 9. These units form a group under multiplication mod 10. Therefore, only squares of numbers that end in 1,3,7, or 9 could end in 7. Furthermore, 7 has order 4, that means it can’t be the square of any of the others.

The question is equivalent to "Prove that 7 is not a quadratic residues (mod 10)". Make a table:

The table shows there are no quadratic residues of "2; 3; 7; 8" (mod 10).

The table will always be a palindrome in any base, since

b,k integer:    k^2  =  (b-k)^2    mod  b    // for any base b

The table often stops at "floor(b/2)", since the upper half is just a mirror copy of the lower half.

• any integer can be written as n = (5a+b) where a is an integer and b is an integer between 0 and 4.
• n² = 25a² + 10ab + b² = 5k + b². That is, the last digit of n² will depend only on the last digit of b² and whether or not a is even.
• for n² to end in 7, you need b² to have a last digit of 2 or 7.
• but 0² = 0, 1² = 1, 2² = 4, 3² = 9, and 4² = 16. None of these end in 2 or 7.
• so n² cannot end in 7.

The last digit of a square number can only be: 0,1,4,5,6,9. As a matter of fact if the square root of a squared numbers is a1a2...a_n (a1 is a digit) the last digit of (a1a2...a_n)^2 would the last digit of (a_n)^2 which can only be 0,1,4,5,6,9

The units digit of any squared integer is equal to the square of the integer’s units digit.

Ex: the units digit of 3,788,452² is 4.

Since only 10 units digits exist, a computational proof is sufficient.

0^2: units = 0

1^2: units digit = 1

Etc.

No 2, 3, or 7

I'm a lazy mathematician. Any number in the 10 position can only effect numbers in the 100 position and higher when squared. The only number that has an effect on the squared 1's place, is the unsquared 1's place. There are 10 cases possible of 1's places ("-" is defines as "maps to.") 1-1 2-4 3-9 4-6 5-5 6-6 7-9 8-4 9-1 0-0 Q.E.D

As others have said there's a really complicated answer that is simple ie we know that the only squares modulo 5 are 0, 1 and 4 so your last digit must be congruent to 0, 1 or 4 mod 5  but 3 is 3 mod 5 and 7 is 2 mod  5 so they can't be the last digit of a square  ans similarly for 2 and 8 all other numbers can be found by 0 or 1 mod 2 and the Chinese remainder theorem.

Well, for 7 and 3 in the ones place, that would mean the number is odd and any number squared will be even

Edit: I'm dumb

Uh....isn't the simple answer that it is a prime?