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Prove ${\forall x, y \in \mathbb{N}}$ there does not exist $x, y$ s.t. $y^ 2 = x^2 + 2 x y$ without using $\sqrt{2}$.

Proving the statement with $\sqrt{2}$ is just a matter of taking the square-root of both sides:

$$\Longrightarrow y^2 = x^2 + 2xy$$ $$\Longrightarrow 2y^2 = (x+y)^2$$ $$\Longrightarrow y\sqrt{2} = (x+y)$$

Since $\sqrt{2}$ is irrational, and $x+y$ is a natural number, there is a contradiction.

There is also a geometric aid for this (the binomial expansion), there is a missing $y^2$. But how can I prove this algebraically?

I tried to factor $x^2 + 2xy$ into $x(x + 2y)$, and assuming that $x$ and $x + 2y$ are perfect squares, but that reasoning may be too specific. I tried rearranging the terms and finding a contradiction but nothing works. Any guidance or answer will be greatly appreciated.

David G. Stork
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MaximeJaccon
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2 Answers2

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Use induction on $x+y$, clearly there do not exist $x',y' \in \mathbb{N}$ with $x'+y'=2$ such that $2(y')^2=(x'+y')^2$. So we assume the following (IH): for all $x',y' \in \mathbb{N}$ with $x'+y' < x+y$, the relation $2(y')^2 \not = (x'+y')^2$ holds.

  1. Suppose that $y=1$. Then the inequality $x<y$ must also hold, since $y^2$ is $x^2+2x$, with $x>0$ and thus $y^2 =x^2 +$ [something positive], which gives $y^2>x^2$ which gives [since $y,x$ positive] $y>x$. But as we are given $x \ge 1$ this is impossible if $y=1$.

  2. So in light of 1. above: Let $p$ be a prime that divides $y^2$; this $p$ exists as $y$ is an integer at least $2$. Then $p$ must divide both $y$ and the other side $(x+y)^2$ of the equation $2y^2=(x+y)^2$, which in turn implies that $p$ divides both $y$ and $(x+y)$, which implies that $p$ divides $x$ as well, which gives $\frac{x}{p}$, $\frac{y}{p}$ both integral. However, dividing both sides of the equation $2y^2=(x+y)^2$ by $p^2$ gives

$$2\Big(\frac{y}{p}\Big)^2 = \Big(\frac{x}{p}+\frac{y}{p}\Big)^2,$$ with $\frac{x}{p}$, $\frac{y}{p}$ each integral [and positive]. This however contradicts (IH), as $x'=\frac{x}{p}$, $y'=\frac{y}{p}$ are integral and positive and thus in $\mathbb{N}$, with $x'+y' < x+y$. And so we are done.

Mike
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I regard the analysis in the original posting as valid, assuming that the assertion is that there are no non-zero integers $~x,y~$ such that $~y^2 = x^2 + xy.~$

That is, $~\sqrt{2} \times y~$ must be an irrational number.

Alternative (somewhat similar) approach:

The following analysis indicates that you can not have $~y~$ be any element in $~\Bbb{Z_{\geq 2}}.$

Given $~n \in \Bbb{Z_{\geq 2}},~$ with prime factorization shown in (1) below, then $~n~$ is a perfect square if and only if each element in $~\{a_1, a_2, \cdots, a_r\}~$ is even:

$$n = p_1^{a_1} \times p_2^{a_2} \times p_r^{a_r}. \tag1 $$

Then

$$y^2 = x^2 + 2xy \iff 2y^2 = x^2 + 2xy + y^2 = (x+y)^2. \tag2 $$

However, the conclusion reached in (2) is contradicted by the analysis this answer's first paragraph. That is, every exponent involved in the prime factorization of $~y^2,~$ must be even. Therefore, the exponent assigned to the prime number $~2~$ in the prime factorization of $~(2y^2)~$ must be odd. This contradicts the assertion that $~(2y^2) = (x+y)^2,~$ a perfect square.

By similar analysis, $~y=1~$ must also be rejected, since then, $~2y^2 = 2,~$ which is also not a perfect square.

In a sense, you could intuit that $~y = 0~$ permits a solution because in the equation $~(x+ 0)^2 = 2^1 \times 0^2 = 0, ~$ it is as if the factor $~2^1~$ disappears.

user2661923
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