# Number TheoryProof review: x² - y² = 270

#### sweatingbear

##### Member
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 $$\displaystyle x$$ and $$\displaystyle y$$ such that $$\displaystyle x^2 -y^2 = 270$$.

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

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

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

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

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

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

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

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

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

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

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

Any thoughts?

#### Opalg

##### MHB Oldtimer
Staff member
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.

##### Well-known member
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 $$\displaystyle x$$ and $$\displaystyle y$$ such that $$\displaystyle x^2 -y^2 = 270$$.

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

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

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

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

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

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

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

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

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

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

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

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

##### Member
Fantastic, thank you!

#### Sudharaka

##### Well-known member
MHB Math Helper
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.