MHB Is this proof of x² - y² = 270 being unsolvable valid?

[SweatingBear]

Forum, I would love if you had a look at my proof below and gave me some feedback. Is the approach valid? Perhaps you have an alternative way to solve the problem? Anything helps!

Problem: Prove that there exist no positive integers $$x$$ and $$y$$ such that $$x^2 -y^2 = 270$$.

Proof: Given that $$x, y \in \mathbb{N}$$ where $$\mathbb{N} = \{ 1, 2, 3, \ldots \}$$, we can infer that $$x$$ and $$y$$ respectively will be either even or odd. That means we have four different cases to examine.

Case #1: $$x$$ even, $$y$$ even. Let $$x = 2k $$ and $$y = 2p$$ ($$k,p\in\mathbb{N}$$); this yields

$$(2k)^2 - (2p)^2 = 270 \iff 4k^2 - 4p^2 = 270 \iff k^2 - p^2 = \frac {270}4 \, .$$

Note now that $$k^2 - p^2$$ itself is in an integer but $$\frac {270}4$$ is not, which clearly is nonsensical. $$k$$ and $$p$$ can therefore not be integers and consequently neither $$x$$ and $$y$$ (at least not even integers).

Case #2: $$x$$ even, $$y$$ odd. Let $$x = 2k $$ and $$y = 2p-1$$ ($$k,p\in\mathbb{N}$$); this yields

$$(2k)^2 - (2p-1)^2 = 270 \iff 4(k^2 - p^2 + p) - 1 = 270 \iff k^2 - p^2 + p = \frac {271}4 \, .$$

Similarly, $$k^2 - p^2 + p$$ is integral but $$\frac {271}4$$ is not; nonsensical of course and thus $$k$$ and $$p$$ cannot be integral (and consequently neither $$x$$ and $$y$$).

Note: the case $$x$$ odd and $$y$$ even is analogous and will therefore be skipped.

Case #3: $$x$$ odd, $$y$$ odd. Let $$x = 2k - 1 $$ and $$y = 2p-1$$ ($$k,p\in\mathbb{N}$$); this yields

$$4(k^2 - k - p^2 + p) = 270 \iff k^2 - k - p^2 + p = \frac {270}4 \, .$$

Again, this leads us to the conclusion that $$k$$ and $$p$$ cannot be integers and therefore neither $$x$$ and $$y$$. $$\text{Q.E.D.}$$

Any thoughts?

 

[Opalg]

That argument looks fine. The key to it is the fact that 270 is a multiple of 2 but not a multiple of 4. In fact, if $x^2-y^2 = (x+y)(x-y) = 270$ then one of the factors $x+y$, $x-y$ must be even and the other one odd. But $x+y = (x-y) + 2y$, so $x+y$ and $x-y$ have the same parity (either both even or both odd), so their product cannot be 270.

 

[kaliprasad]

Forum, I would love if you had a look at my proof below and gave me some feedback. Is the approach valid? Perhaps you have an alternative way to solve the problem? Anything helps!

Problem: Prove that there exist no positive integers $$x$$ and $$y$$ such that $$x^2 -y^2 = 270$$.

Proof: Given that $$x, y \in \mathbb{N}$$ where $$\mathbb{N} = \{ 1, 2, 3, \ldots \}$$, we can infer that $$x$$ and $$y$$ respectively will be either even or odd. That means we have four different cases to examine.

Case #1: $$x$$ even, $$y$$ even. Let $$x = 2k $$ and $$y = 2p$$ ($$k,p\in\mathbb{N}$$); this yields

$$(2k)^2 - (2p)^2 = 270 \iff 4k^2 - 4p^2 = 270 \iff k^2 - p^2 = \frac {270}4 \, .$$

Note now that $$k^2 - p^2$$ itself is in an integer but $$\frac {270}4$$ is not, which clearly is nonsensical. $$k$$ and $$p$$ can therefore not be integers and consequently neither $$x$$ and $$y$$ (at least not even integers).

Case #2: $$x$$ even, $$y$$ odd. Let $$x = 2k $$ and $$y = 2p-1$$ ($$k,p\in\mathbb{N}$$); this yields

$$(2k)^2 - (2p-1)^2 = 270 \iff 4(k^2 - p^2 + p) - 1 = 270 \iff k^2 - p^2 + p = \frac {271}4 \, .$$

Similarly, $$k^2 - p^2 + p$$ is integral but $$\frac {271}4$$ is not; nonsensical of course and thus $$k$$ and $$p$$ cannot be integral (and consequently neither $$x$$ and $$y$$).

Note: the case $$x$$ odd and $$y$$ even is analogous and will therefore be skipped.

Case #3: $$x$$ odd, $$y$$ odd. Let $$x = 2k - 1 $$ and $$y = 2p-1$$ ($$k,p\in\mathbb{N}$$); this yields

$$4(k^2 - k - p^2 + p) = 270 \iff k^2 - k - p^2 + p = \frac {270}4 \, .$$

Again, this leads us to the conclusion that $$k$$ and $$p$$ cannot be integers and therefore neither $$x$$ and $$y$$. $$\text{Q.E.D.}$$

Any thoughts?

it looks good

a shorter proof shall be

x^2 - y^2 = (x-y)(x+y) = (x- y)(x- y+ 2y)

either both are odd or even so product is odd or multiple of 4 if even,
270 is multiple of 2 and not 4 and hence not possible

I am sorry. I had not seen Oplag's proof which is almost same.

 

[SweatingBear]

Fantastic, thank you!

 

[Sudharaka]

Hi everyone, :)

Another method to prove this is to use the fact that every square integer when divided by \(4\) gives a remainder of \(0\) or \(1\). That is, \(x^2\equiv n\mbox{(mod 4)}\) has solutions if and only if \(n\equiv 0,\,1\mbox{(mod 4)}\). Hence the remainder of \(x^2-y^2\) divided by \(4\) should be either \(0,\, 1\) or \(3\). But the remainder of \(270\) divided by \(4\) is \(2\). Hence we arrive at the conclusion that \(x^2-y^2=270\) has no integer solutions.