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Perfect Squares with Congruences


Date: 02/16/2002 at 14:17:15
From: Stacey
Subject: Perfect squares with congruences

Prove that there is no perfect square a^2 whose last digits are 35.


Date: 02/17/2002 at 17:42:47
From: Doctor Paul
Subject: Re: Perfect squares with congruences

What if I asked you to prove that there is no perfect square a^2 whose 
last digit is 7?

Notice that every integer falls into one of 10 classes:

   those that are 

     multiples of 10
     1 more than a multiple of ten
     2 more than a multiple of ten
     3 more than a multiple of ten
     4 more than a multiple of ten
     5 more than a multiple of ten
     6 more than a multiple of ten
     7 more than a multiple of ten
     8 more than a multiple of ten
     9 more than a multiple of ten

Written in terms of moduli, we can say that every integer is either 
congruent to 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 mod 10.

Notice that the last digit of a^2 is just the same thing as computing 
a^2 mod 10. So we do this for the 10 possibilities:

If an integer is congruent to 0 mod 10, then x^2 = 0^2 = 0 mod 10

   i.e., if x = 0 mod 10 then the last digit of x^2 is 0.

If an integer is congruent to 1 mod 10, then x^2 = 1^2 = 1 mod 10

   i.e., if x = 1 mod 10 then the last digit of x^2 is 1.

If an integer is congruent to 2 mod 10, then x^2 = 2^2 = 4 mod 10

   i.e., if x = 2 mod 10 then the last digit of x^2 is 4.

If an integer is congruent to 3 mod 10, then x^2 = 3^2 = 9 mod 10

   i.e., if x = 3 mod 10 then the last digit of x^2 is 9.

If an integer is congruent to 4 mod 10, then x^2 = 4^2 = 16 = 6 mod 10

   i.e., if x = 4 mod 10 then the last digit of x^2 is 6.
.
.  [fill in the details]
.

If an integer is congruent to 9 mod 10, then x^2 = 9^2 = 81 = 1 mod 10

   i.e., if x = 9 mod 10 then the last digit of x^2 is 1.

This shows that no square has 7 as its last digit.

A similar argument holds for showing that the last two digits of a 
number will never be 35. Just square each of the 100 residue classes 
mod 100 and show that none of them yields 35. It's computationally 
time-consuming so I wrote a little Maple script to do it for me:

> restart;
> L := [seq(x[i],i=1..100)]:
> for i from 1 to 100 do
> L[i]:=i^2 mod 100;
> od:
> L;

  [1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89, 24,

        61, 0, 41, 84, 29, 76, 25, 76, 29, 84, 41, 0, 61, 24, 89, 56,

        25, 96, 69, 44, 21, 0, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0, 1,

        4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89,

        24, 61, 0, 41, 84, 29, 76, 25, 76, 29, 84, 41, 0, 61, 24, 89,

        56, 25, 96, 69, 44, 21, 0, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0

        ]

Inasmuch as none of the squares mod 100 is 35, we can conclude that 
there is no number whose square has 35 as its last two digits.

I hope this helps.  Please write back if you'd like to talk about this 
more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory

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