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Number Theory Proof review: x² - y² = 270

sweatingbear

Member
May 3, 2013
91
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
Feb 7, 2012
2,707
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

Well-known member
Mar 31, 2013
1,309
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
May 3, 2013
91
Fantastic, thank you!
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.