Squares and Squares of Integers

Relevant wiki: Quadratic Diophantine Equations - Fermat's Method of Infinite Descent

Proof by contradiction

Take the positive integer solution tuple $x,y,z$ such that $z$ is the smallest possible.

Because $1^2\equiv2^2\equiv1\ (\text{mod }3)$ and the right-hand side of the equation is divisible by $3$ , it must be the case that both $x$ and $y$ are divisible by $3$ .

So $x=3a$ and $y=3b$ for some positive integers $a$ and $b$ .

Then, $9a^2+9b^2=3z^2$ , or $3a^2+3b^2=z^2$ . This means that $z$ is also divisible by $3$ . So $z=3c$ for some integer $c$ .

This means that $9a^2+9b^2=27c^2$ .

Divide both sides by $9$ to get $a^2+b^2=3c^2$ . Since $a<x,b<y,c<z$ , this means that there is a smaller positive integer solution than $x,y,z$ , contradicting our original assumption.

This proves that there is no smallest positive integer solution, or no positive integer solution at all. The argument works similarly for negative numbers. In other words, since it is impossible for exactly one or two of the variables to be $0$ , $x=y=z=0$ is the only possible integer solution.

Relevant wiki: Fermat's Method of Infinite Descent

Consider the remainder of $x^2, y^2$ when divided by $3$ , it can be only $0$ or $1$ . Note that $x^2+y^2$ is a multiple of $3$ . If either one of remainder when $x^2$ or $y^2$ is divided by $3$ is equal to $1$ , then the other remainder should be $2$ , which is impossible. Hence, we can conclude that $x^2$ and $y^2$ are multiple of $3$ , which is also $x$ and $y$ .

Let $x=3a, y=3b$ . We can get $\begin{aligned} x^2+y^2&=3z^2 \\ 9a^2+9b^2&=3z^2 \\ 3a^2+3b^2&=z^2 \\ 3(a^2+b^2)&=z^2 \end{aligned}$ Obviously, $z$ is multiple of $3$ . Let $z=3c$ . $\begin{aligned} 3a^2+3b^2&=9c^2 \\ a^2+b^2=3c^2 \end{aligned}$ Now, we get another equation with same style but different numbers. Obviously, $a,b,c$ are smaller than $x,y,z$ respectively.

Let there exist another non-trivial solution ( $x_0,y_0,z_0$ ) then we can get another $smaller$ non-trivial solution ( $a_0,b_0,c_0$ ). And we can get another smaller non-trivial solution again infinitely. However, this is obviously non logical, because there exist finite number which are smaller than $x_0$ . Hence, by contradiction, there is $\large no$ non-trivial solution.

It is a well-known fact that the sum of two coprime perfect squares is the product of 2 and/or prime numbers of the form $4n+1$ . It can therefore not be divisible by 3.

This one is also through fermat's method of infinite descent (with a twist).
Instead of modulo 3 go modulo 4.
The set of quadratic residues modulo 4 is {0,1} The only permissible case for x,y,z is $x²,y²,z² \equiv 0\pmod 4$ .
If they leave any of the remainder the equation is not satisfied. So, $x,y,z \equiv 0\pmod 2$ .
$Let x=2a, y=2b, z=2c$ .
Substituting and simplifying we get,.
$a²+b²=3c²$ .
Now I think you know how to apply fermats method of infinite descent and show that the given equation has no positive integer solutions.
Again as the terms are squares, so as there are no positive integer solutions therefore there are no integer(non-trivial) solutions.

The prime decomposition of $3z^2$ will necessarily contain 3 raised to an odd power, when $z \neq 0$ . Therefore, by the sum of two squares theorem , $3z^2$ cannot be expressed as a sum of integer squares, when $z \neq 0$ .

I do not expect any upvotes for this, totally lacking in style.

The right hand side is divisible by 3 and must have an odd number of factors of 3 (3 plus an even number of 3's from z squared).

First, if 3 divides x, it must also divide y otherwise the left hand side will not be divisible by 3. If both x and y are divisible by 3, the sum of their squares will have an even number of factors of 3 and thus cannot equal the product of an odd number of factors on the right hand side.

Second, the squares of all numbers not divisible by 3 have remainder 1 when divided by 3. Thus, if neither x nor y is divisible by 3, the left had side will have remainder 2, while the right hand side is divisible by 3 with no remainder.

That is how I solved it.

If $(a,b,c)$ is a primitive Pythagorean triplet then exactly one of $a,b$ is divisible by 3. The given equation implies both $a,b$ is divisible by 3. A contradiction if they are non-zero.

$x^2 +y^2 = 3z^2$

$x + y = \sqrt(3) z$

Now we know that x + y divided by the square root of three is z. This means that of x and y are integers. Z is always irrational and visa versa because a divided by b where b is irrational and a is rational is an irrational number

Deploying primitive Pythagorean triplets here ( $p,q \in \mathbb{N}, p > q)$ yields:

$x = p^2 - q^2$

$y = 2pq$

$\sqrt{3} z= p^2 + q^2$

If $z = \frac{p^2+q^2}{\sqrt{3}},$ then $z$ is purely irrational $\Rightarrow z \notin \mathbb{Z}.$ Hence, no other integral triplet than $x=y=z=0$ satisfies $x^2 + y^2 = 3z^2.$

Note that $x^2 + y^2 \equiv 0 \ (mod \ 3)$ . But then by Fermat's little theorem we have $1+1 \equiv 0 \ (mod \ 3)$ which is a contradiction and leads us to the conclusion that there are no more integer solutions.

You have that x=y=z so u can all replace them by one variable and there you go.

x^2+x^2=3x^2

2x^2 /x^2 =3

2=3 impossible

If you subtract, it will lead you to the old statment that x=y=z=0

You can think it logically also. A equation in 2 variable generally have infinitely many solutions but as you see it is in 3 variables.

The key is $x^2 \equiv1\ \text{or}\ 0 (\text{mod}\ 3)$

Here's a different way to do it:

The equation defines a surface in $\mathbb{R}3$ . Express that surface in cylindrical coordinates: $(r sin(\theta))^2 + (r cos(\theta))^2 = 3 z^2 = r^2$

Take the square root: $r = \sqrt{3} z$

Since $\sqrt{3}$ is irrational, $\frac{r}{z}$ cannot equal $\sqrt{3}$ for integer values of r and z.

I followed an entirely different part, so I wanted to check this with you guys (and try LaTeX for the first time :) ).

We start from $x^2 + y^2 = 3z^2$
Assume that $y \geq x$ , so $y = k \cdot x$ , with $k \geq 1, k \in \mathbb{R}$ . If $y < x$ , switch them.

Then $x^2 + y^2 = x^2 + k^2 \cdot x ^2 = x^2 \cdot (k^2 + 1) = 3 \cdot z^2$
Therefore, $z^2 = x^2 \cdot \frac {k^2 + 1}{3}$

Note by definition $z \geq x$ , and $z \geq y$ , so $z \geq k \cdot x$
Then $z^2 = x^2 \cdot \frac {k^2 + 1}{3} \geq x^2 \cdot k^2$

This means either $x = 0$ (and $y = 0, z = 0$ ) or $k^2 + 1 > 3 \cdot k^2$
The latter means $1 > 2 \cdot k^2$ , or $k^2 < \frac{1}{2}$ .

This has only solutions for $- \frac{1}{2} \sqrt{2} \leq k \leq \frac{1}{2} \sqrt{2}$ , but we have already defined $k \geq 1$ .
Therefore, no solution exists other than $x = y = z = 0$ .