Question

\(\large \begin{align} \color{black}{\normalsize \text{prove that the equation has no integer solution}\hspace{.33em}\\~\\

x^2+y^2=57\hspace{.33em}\\~\\}\end{align}\)

Answer 1

well, hmm let's see

Answer 2

So do you agree that if we restrict ourselves to integers only the x^3 could ever be negative?

Answer 3

btw, I'm still not sure I buy this claim

Answer 4

When x is positive, we know that both x and y are odd at the very least. The hardest thing to show here looks like when y is really large and x is really large and negative.

Answer 5

yea, That's what I was thinking about Kainui, do you think if we did some form of mod or common divisor thing we could exclude large values of each?

Answer 6

Beats me honestly, I'm sorta just stumped trying out random stuff.

Answer 7

And I think we could quickly eliminate all positive x,y

Answer 8

It might be nice to rewrite x^3 as -x^3 since it is arbitrary.

Answer 9

we'd only need the cubes less than 56 and squares less than 56 to eliminate positive combos

Answer 10

Yeah we can easily exhaust our way through the possibilities there, and if we're proving it I mean we can just do that last since it's straight forward, I am imagining that will work out simply enough but idk. \[\large y^2 = 57+x^3\] That's kind of what I'm looking at atm, idk.

Answer 11

well, what if we reduced this to mod19. \(\frac{57}{3}=19\)

Answer 12

we would be able to look at only x,y=1,2,3...19. Set's of combinations

Answer 13

or mod 3 for that matter..

Answer 14

so we assume that \(x^3+y^2\equiv 0 mod 3\)

Answer 15

or we assume \(x^3+y^2\equiv 0 mod 19\)

Answer 16

We just made two equations with two unknowns? @Kainui what do you think?

Answer 17

I am not really all that good at mods honestly so I don't know what to think, it seems like you're doing something though haha.

Answer 18

I would probably do mod 3 since it looks like that means we have less to check?

Answer 19

but that would be an issue if they were both 0 mod 3?

Answer 20

grr

Answer 21

\[\Large (3n+a)^3 + (3m+b)^2 \equiv 0 \mod 3 \\ \Large a^3 +b^2 \equiv 0 \mod 3\] But now we're looking at a and b only being 0, 1 or 2 so we can rule out some stuff. Maybe we can then do the same thing with mod 19 and look at the intersection of these two sets?

Answer 22

maybe like chinese remainder hmm...

Answer 23

@mathmath333 what class is this for?

Answer 24

I guess a simpler way just to look at it is to graph it as \[y=+-(57-x^3)^{1/2}\]

Answer 25

that should only have at max 3 zeroes right?

Answer 26

Ooooh that seems like that method will work actually.

Answer 27

but that's not a "proof"

Answer 28

that's my only issue

Answer 29

but anyways,im annoyed that the guy who asked the question isn't even online so I'm gonna stop

Answer 30

Well I mean how is it not a proof?

Answer 31

a graph is not a proof

Answer 32

Unless it crosses at an integer, which are easily checked and we can tell if the function is always increasing with calculus so we know it won't cross again, that kind of thing, seems like it is proof.

Answer 33

I think you are wrong about what is and isn't a proof.

Answer 34

^ that is not a proof either

Answer 35

This isn't much different than the IVT.

Answer 36

and no, trust me, I have been marked incorrect enough times to know this is not a proof

Answer 37

we would have to write this up in a rigorous way using the ivt

Answer 38

and some other stuff

Answer 39

Being marked incorrect is not a logical argument, it's a logical fallacy, appealing to authority.

Answer 40

no, being told it is not a proof in analysis courses has made me of the notion, that it isn't

Answer 41

Perhaps your teacher expected you to show something was true through a certain method that you were being tested on, which is different. There's no reason to say that this is illegitimate when it shows what is true.

Answer 42

it is not rigorous, hence not a proof. They(multiple professors) don't care what method, it must be rigorous, and this is not.

Answer 43

My math department is very very analysis heavy

Answer 44

What is not rigorous about it?

Answer 45

The fact that we are using a picture one. 2, no formal write up, 3. still not proved

Answer 46

I mean if you do a formal ivt write up, maybe

Answer 47

There is no need for a picture, and even if we were, you are basically slapping Euclid and saying he has no rigor. It can be formally written up.

Answer 48

but we haven't done that

Answer 49

(but we haven't formally written it up, hence not proof)

Answer 50

Hahaha we don't have to, that's @mathmath333 's job, we have found the way to the answer though, that's all I am worried about.

Answer 51

lol

Answer 52

Don't get me wrong though, a geometric proof is still a proof.

Answer 53

\(\large\tt \begin{align} \color{black}{\normalsize \text{lol counter example }\hspace{.33em}\\~\\

2^3+7^2=57\hspace{.33em}\\~\\}\end{align}\)

Answer 54

sry guys the question is this one
\(x^2+y^2=57\)