Prove Square of Integer Ends in 0, 1, 4, 5, 6, 9

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SUMMARY

The square of an integer in base 10 can only end in the digits 0, 1, 4, 5, 6, or 9. This conclusion is derived by analyzing the last digit of integers from 0 to 9 and their squares. The proof involves expressing any integer as n = 10i + j, where j represents the last digit, and calculating the square for each possible last digit. The results confirm that only the specified digits can occur as the last digit of a square.

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How can i prove that square of an integer ends with 0,1,4,5,6,9 ?
 
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Martin Bashir said:
How can i prove that square of an integer ends with 0,1,4,5,6,9 ?

What base is your number system in ? If it is ten then you have to show that with all ten possible endings, n squared will not give a number with a different ending.
As another thought since n squared = -n squared in any base the maximum number of possible endings is [(B + 1)/2] where [] is the next higher integer and B is the base. In base 10 you will only need to look at numbers from 0 to 10/2 to find the possible endings since 5 to10 are the same as - (-5 to 0) mod 10.
 
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You can write any number, in base 10, as n= 10i+ j for integers i and j with j the "last digit", from 0 to 9.
[itex]10i+ 0= 100i^2[/itex] which ends with 0.
[itex]10i+ 1= 100i^2 + 20i+ 1= 10(10i^2+ i)+ 1[/itex] which ends with 1.
[itex]10i+ 2= 100i^2+ 40i+ 4= 10(10i^2+ 4i)+ 4[/itex] which ends with 4.
[itex]10i+ 3= 100i^2+ 60i+ 9= 10(10i^2+ 6i)+ 9[/itex] which ends with 9.
[itex]10i+ 4= 100i^2+ 80i+ 16= 10(10i^2+ 8i+ 1)+ 6[/itex] which ends with 6.
[itex]10i+ 5= 100i^2+ 100i+ 25= 10(10i^2+ 10i+ 2)+ 5[/itex] which ends with 5.
[itex]10i+ 6= 100i^2+ 120i+ 36= 10(10i^2+ 12i+ 3)+ 6[/itex] which ends with 6.
[itex]10i+ 7= 100i^2+ 140i+ 49= 10(10i^2+ 14i+ 4)+ 9[/itex] which ends with 9.
[itex]10i+ 8= 100i^2+ 160i+ 64= 10(10i^2+ 16i+ 6)+ 4[/itex] which ends with 4.
[itex]10i+ 9= 100i^2+ 180i+ 81= 10(10i^2+ 18i+ 8)+ 1[/itex] which ends with 1.
 

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