It's called a perfect square every natural number can be written as a power of two other natural number.

**a perfect square ‹=› a = k ^{2}**

It's called perfect cube that any natural number can be written as the third power of another natural number.

**a perfect cub ‹=› a = k ^{3}**

Here are some known results in working with power:

** 1) **The last digit of a perfect square is just one of the following

**0, 1, 4, 5, 6, 9**.

** 2) **Any perfect square is one of the forms

(Indeed, if *n *= 2*k*, then *n*^{2} = 4*k*^{2} = 4*p*, and if *n *= 2*k *+1, we *n*^{2} = 4*k*(*k *+1) +1= 8*q *+1)

** 3) **Any perfect square is one the forms 3

(As above, we consider *n *= 3*k *or *n *= 3*k *+1 or *n *= 3*k *+ 2 and we raise to square)

** 4) **If a perfect square contains a prime factor decomposition, this factor is actually in power seem to decompose the initial number.

** 5) **The rest of dividing any perfect square through 4 is 0 or 1.

** 6) **A number that ends in one of the figures 2, 3, 7, or 8 is not a perfect square.

** 7) **To show that a number is not a perfect square we can show that he is between two squares of consecutive numbers.

**How many natural numbers, integers between two consecutive perfect squares?**

Natural numbers are consecutive perfect squares:

0^{2}; 1^{2}; 2^{2}; 3^{2}; 4^{2}; 5^{2}; 6^{2}; 7^{2}; 8^{2}; 9^{2}; 10^{2}; ...,or, more precise:

0; 1; 4; 9; 16; 25; 36; 49; 64; 81; 100; ...

Between 0 and 1 are 1-0-1=0 natural numbers;

Between 1 and 4 are 4-1-1=2 natural numbers;

Between 4 and 9 are 9-4-1=4 natural numbers;

Between 9 and 16 are 16-9-1=6 natural numbers;

Between 16 and 25 are 25-16-1=8 natural numbers;

Between 25 and 36 are 36-25-1=10 natural numbers;

Between 36 and 49 are 49-36-1=12 natural numbers;

It notes that the results are numbered consecutively. We pick up again what I wrote above except that we write perfect squares as power exponent 2:

Between 0^{2} and 1^{2} are 0 natural numbers, i.e. 2·0;

Between 1^{2} and 2^{2} are 2 natural numbers, i.e. 2·1;

Between 2^{2} and 3^{2} are 4 natural numbers, i.e. 2·2;

Between 3^{2} and 4^{2} are 6 natural numbers, i.e. 2·3;

Between 4^{2} and 5^{2} are 8 natural numbers, i.e. 2·4;

Between 5^{2} and 6^{2} are 10 natural numbers, i.e. 2·5;

Between 6^{2} and 7^{2} are 12 natural numbers, i.e. 2·6;

Generalizing, we can say that between n^{2} and (n+1)^{2} exists 2·n natural numbers, where „n” is any natural number.

**Exercise 1**

Learn two perfect square integers are integers knowing consecutive perfect squares and among them there are 60 integers.

**Solve**

Let be n^{2} and (n+1)^{2} the two consecutive perfect square integers, where

* n* is a some natural number.

We have the relation: 2·n=60 => n=30.

So the two consecutive integers that are perfect squares 302 and 312.

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