Question

Let X and Y be natural numbers.
Prove that x^2+y+2 and y^2 +4x cannot be both perfect squares.
Please help

Answer 1

I do not have an idea to start this

Answer 2

hmmm, it says nothing about the same perfect square number does it

Answer 3

I guess not

Answer 4

what class is this for?

Answer 5

This is a maths Olympiad question (1990 I think)

Answer 6

lets assume they are and prove that its not

Answer 7

is the sum of perfect squares a perfect square? what property do perfect squares have that we can demonstarte?

Answer 8

9+4 = 13 so the sum isnt

Answer 9

I tried to do somthing with a^2-b^2= (a-b)(a+b) but got nowhere

Answer 10

a^2 + b^2 = c^2 is the sum of perfect squares as a perfect square right?

Answer 11

Nope, that is Fermat's little theorem. Right?

Answer 12

Or that is with cubes? :) Was long time ago

Answer 13

i was thinking pythag thrm, but yeah.

Answer 14

fermats is a^n + b^n = c^n for n>2

Answer 15

Oh yeah :) Never mind, that is offtopic now

Answer 16

start with a if p then q statement

Answer 17

see if the contrapositive is easier to determine, maybe

Answer 18

let x^2+y+2 = a^2
let y^2+4x = b^2


x^2+y+2 + y^2+4x = c^2

(x^2+4x)+ (y^2+y) = c^2 -2

(x^2+4x+4)+ (y^2+y+1/4) = c^2 -2+4+1/4

(x+2)^2+ (y+1/2)^2 = c^2 + 2.25

dunno if this gets us anywhere, but its about the only thought i got at the moment

Answer 19

x^2+y+2 + y^2+4x = c^2 This is not necessarily true. Only for Pythagoras triplets

Answer 20

i agree, but can we define x and y so that it make a pythag triple?

Answer 21

That is not given, so I guess we cannot.

Answer 22

we just need to prove it false, so one counter example is all we need

Answer 23

No because it says that we only need one good X, Y (anything)

Answer 24

then for a start, think about properties of perfect squares then apply that property to what we have to see if it holds

Answer 25

pythag has perfect squares so i thought it might be prudent. im open to being wrong of course

Answer 26

i remember working this problem the other day in mse

Answer 27

Well thanks for trying, I have to go now. Will think on it tomorrow. It bothers me though

Answer 28

i dont remember working it at all in over 40 years :)

Answer 29

hold up, it is easy

Answer 30

maybe: a^2/b^2 = (a/b)^2 ?

Answer 31

I found the problem: http://www.imomath.com/index.php?options=Hi&mod=23&ttn=Hungary-Israel
No solutions there...

Answer 32

Smart

Answer 33

some property of perfect squares :) just needed to determine the property

Answer 34

does it matter than not all k=0,1 (mod4) are perfect squares?

Answer 35

ill trust you on that for now lol

Answer 36

@amistre64 i see my stupid mistake, it wont work yeah :/
will get back to this after dinner..

Answer 37

see, i trust you to find your way ;)

Answer 38

:) found this
http://artofproblemsolving.com/community/c6h392528_prove_there_are_no