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Show that there is no right triangle whose legs are rational numbers and whose hypotenuse is $\sqrt{2022}$. My tries:

  1. I used Pythagoras' Theorem to get: $$\sqrt{2022}^2=a^2+b^2 \implies a^2+b^2 = 2022$$ where $a$ and $b$ are the legs of the triangle. I don't know what to do next: Is there another formula I could use? I know that $a+b>\sqrt{2022}$ but I don't think this is going to help us much.

hope one of you can help me! thank you!

D S
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  • If you were to rewrite $a$ and $b$ with common denominator $k$, you then need to show that $2022,k^2$ is not the sum of two squares of integers for any positive integer $k$. You might end up using $2022=2 \times 3\times 337$ – Henry Jan 03 '23 at 10:10
  • i dont really understand how you rewrite them with k –  Jan 03 '23 at 10:22
  • Note that $a$ and $b$ are fractions, and then multiply out the denominators of $2022 = a^2 + b^2$. We have a lot more tools for dealing with equations of integers than we do for dealing with equations of rational numbers, so getting rid of all fractions is a quite natural step. – Arthur Jan 03 '23 at 10:24
  • We know $a$ and $b$ are rational. So, we can represent them as $a = \frac xk$ and $b = \frac yk$ for some natural numbers $x$,$y$ and $k$. Then the equation becomes:$2022 = \frac{x^2}{k^2}+\frac{y^2}{k^2} \implies 2022k^2 = x^2+y^2$. – D S Jan 03 '23 at 10:35
  • Pythagoras, not "pitagora". – Dietrich Burde Jan 03 '23 at 10:49
  • Where did this question come from? – Adam Rubinson Jan 03 '23 at 11:23
  • @DietrichBurde Not Pythagoras, it's Πυθαγόρας, as your Wikipedia link clearly states. As for how the OP spells it in their own language, it may very well be "Pitagora", or the closest equivalent to it if they don't use our lettters. And they may not have encountered how his name is usually spelled in English. – Arthur Jan 03 '23 at 11:54
  • @Arthur The title of the wikipedia article says "Pythagoras". I suppose this is the name in English, which is the common language of the posts here. – Dietrich Burde Jan 03 '23 at 12:46

4 Answers4

4

Quickly ruling out $\ a\ $ being even, we try $\ a\ $ and $\ b\ $ both being odd:

$$ (2k_1+1)^2 + (2k_2+1)^2 = 2022\quad k_1,k_2\in\mathbb{Z}$$

$$ \implies 4({k_1}^2 + {k_2}^2 + k_1 + k_2) + 2 = 2022 $$

$$ \implies {k_1}^2 + k_1 + {k_2}^2 + k_2 = 505, $$

which is impossible, since $\ {k_i}^2 + k_i\ $ is even for $\ i=1,2.$

Adam Rubinson
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Let $a=\frac{x}{k}$, $b=\frac{y}{k}$, where $x,y,k$ are positive integers and $\gcd(x,y,k)=1$. Then, the equation becomes $x^2+y^2=2022k^2$. From this equation, it is clear that $x$ and $y$ must have the same parity. If they are both even, then $k$ is odd and the equation has no solution since $4\not|2022$. Therefore, $x$ and $y$ must be odd. If $x,y$ are odd then since square of an odd number modulo $8$ is $1$, we get $k^2\equiv -1\pmod 8$ which is not possible.

Bob Dobbs
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An alternative answer: From the comments, I have shown that $$x^2+y^2 = 2022k^2$$ So, it remains to show that $2022k^2$ can not be represented as the sum of squares of 2 integers. For contradiction, assume that $2022k^2$ can be represented as the sum of $2$ squares. From the Sum of two squares theorem, if $2022k^2$ can be represented as the sum of squares of $2$ integers, it will have prime factors of the form $4n+3$ raised to an even power. $3$ is of the form $4n+3$. Let $$k = 3^pq$$ where $p \ge 0$ and $q$ are integers such that $3$ does not divide $q$. Then, $k^2 = 3^{2p}q^2$. Thus, $$2022k^2 = 2\times 3 \times 337 \times 3^{2p}q^2 = 3^{2p+1} \times 2 \times 337 \times q^2$$We know that $q^2$ does not contain $3$ in its prime factorisation, since $q$ does not contain $3$ in its prime factorisation. So, $2022k^2$ has $3^{2p+1}$ in its prime factorisation and no other $3$. But, $2p+1$ is an odd number. Thus, we get a contradiction, $2022k^2$ cannot be represented as the sum of squares of $2$ integers, meaning $2022$ cannot be represented as the sum of squares of $2$ rational numbers.

D S
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  • Sum of two squares theorem is interesting. – Bob Dobbs Jan 03 '23 at 13:39
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    I came across it (Theorem 3.13) here: http://www.math.uchicago.edu/~may/VIGRE/VIGRE2008/REUPapers/Bhaskar.pdf – D S Jan 03 '23 at 14:06
  • @BobDobbs The more interesting result in my opinion (although irrelevant in this situation) is that the converse holds for primes: If the prime factorization of an integer has an even number of each $4n+3$ prime factor and this integer also is prime (which is to say, if this integer is a prime of the form $4n+1$), then it can be written as the sum of two squares. – Arthur Jan 03 '23 at 14:49
  • I think there is a formula for $r_2(n)$. – Bob Dobbs Jan 03 '23 at 15:01
  • @Arthur primes of the form $4n+1$ can be represented as sum of $2$ squares, as given here: https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares – D S Jan 03 '23 at 16:24
  • see https://encyclopediaofmath.org/wiki/Legendre_theorem I like the treatment in Cassels, Rational Quadratic Forms, pages 80-82, especially the note middle page 82 – Will Jagy Jan 03 '23 at 17:34
  • @DG That is what I said, isn't it? Your comment seems written like you're disagreeing with me. Are you? – Arthur Jan 03 '23 at 20:53
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    @Arthur I'm not disagreeing. Just thought I should add a link to the Wikipedia page for curious readers. – D S Jan 04 '23 at 09:57
1

I think this answer is better than my previous one.

I was doing an unrelated question when I thought of this question and a solution struck me too (don't ask me how). So here it goes:
Write the equation as (using different variables this time): $$\left(\frac pr\right)^2 +\left(\frac qs\right)^2 = 2022$$where these are the fractions in their lowest terms. We get:
$$\color{green}{(ps)^2+(qr)^2 = 2022(rs)^2}$$Both sides are integers, one is divisible by an even power of $3$, the other by an odd power. Hence, there is no solution. (Because in LHS, both terms can only be $0,1 \pmod 3$. Let $ps = 3^mk$ and $qr = 3^nl$ such that $k,l \neq 0 \pmod 3$. Then LHS is divisible by $3^{2m}$ or $3^{2n}$, whichever is lesser. So, LHS is divisible by an even power of $3$. On the other hand, RHS is divisible by an odd power of $3$ since $2022 = 3 \cdot 2 \cdot 337$. )

D S
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