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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.

more.

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

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