Nombres entiers positifs

[alice02]

Déterminez toutes les paires des nombres positifs entiers tels que

est un nombre entier.

[pascal16]

(0;0) is one, maybe not the only one.

[alice02]

no is good. Because the numbers must be

[chan79]

(5;3),(1;3),(2;1),(1;2),(3;1),(2;2),(5;2),(2;5),(3;5)

[alice02]

Yes, but someone can post an analitic solution?
Thanks

[Ben314]

no is good. Because the numbers must be

Be careful : for french mathematicians a "positive" number X mean a number such that .
See for exemple the "bible" of the (french) mathematics (but not only French...), N.Bourbaki where you can read
"0 is the only number witch is both positive and negative".

And if you want to understand why there is such a difference, it's because in France, the only definition we have for a (binary) partial order is "Reflexive + Transitive + Antisymetric" so is a partiel order, but is not a partial order.
In English, there are two definitions : one for "non strict partial order" and one for "strict partial order".

And if I say to be carreful, it's because it's the same problem for other partials orders such as "to be a subset of" : When a french mathematician write , that's always mean .

[alice02]

Ah ok Ben314, I didn't know it.
But someone can post a resolution?
Thanks

[Ben314]

Sinon, pour tout
avec (sinon )
Il faut donc résoudre .
Le discriminant (en ) doit être le carré d'un entier .
- On a évidement vu que .
- Mais d'un autre coté, on a la série d'équivalences suivantes (modulo que ) :
.
Ce qui signifie que, si , la seule possibilité pour est sauf que cette dernière équation conduit (en l'élevant au carré) à clairement sans solution (problème de parité).
Reste à traiter les cas où (c.f. EDIT çi dessous)

P.S. : Il y a surement d'autres méthodes un peu moins "rustres" (par exemple utilisant la divisibilité)...

EDIT :
- Pour y=1, l'unique solution est b=5 (et =1) qui conduit à x=2 ou bien x=3.
- Par symétrie, pour b=1, l'unique solution est y=5 qui conduit à x=2 ou bien x=3.
- Pour b=y=2 l'unique solution (double) est x=2
- Pour b=2; y=3 on a deux solutions x=1 ou bien x=5.
- Par symétrie, pour b=3; y=2 on a aussi deux solutions x=1 ou bien x=5.
Donc sauf erreur 9 solutions : (x,y) { (1;2) ; (1;3) ; (2;1) ; (2;2) ; (2;5) ; (3;1) ; (3;5) ; (5;2) ; (5;3) }

[alice02]

I don't understand all passages...
Can you traslate in english?
Thanks

[Ben314]

Lostounet, il est très fort en anglais...

[alice02]

Ok.

[Lostounet]

Lostounet, il est très fort en anglais...

Translating Ben's post:

Okay, other than that, for all
with (otherwise)

We therefore have to solve the quadratic equation .
Computing its discriminant, we get: which has to be the square of some integer .
- Obviously since .
- On the other hand, we have the following series of logical equivalences (granted that ) :
.
Which means that, if , the only possibility for is however, squaring both sides of the last equation yields which clearly has no solutions (parity incompatibility).
The only cases left to study are (look below)

P.S. : I'm sure there are other less "crude" methods (using divisibility properties of the integers for example)

EDIT :
- For y=1, the only solution is b=5 (and =1) which leads to x=2 or x=3.
- By symmetry, for b=1, the only solution is y=5 which leads us to x=2 or x=3.
- For b=y=2, the only (repeated) solution is x=2
- Take b=2; y=3 we have two solutions x=1 or x=5.
- By symmetry, for b=3; y=2 we also have two solutions x=1 or x=5.
So basically we have 9 solutions : (x,y) { (1;2) ; (1;3) ; (2;1) ; (2;2) ; (2;5) ; (3;1) ; (3;5) ; (5;2) ; (5;3) }

[Ben314]

Merci Lostounet....

[infernaleur]

Lostounet, il est très fort en anglais...

Translating Ben's post:

Okay, other than that, for all
with (otherwise)

We therefore have to solve the quadratic equation .
Computing its discriminant, we get: (en ) which has to be the square of some integer .
- Obviously since .
- On the other hand, we have the following series of logical equivalences (granted that ) :
.
Which means that, if , the only possibility for is however, squaring both sides of the last equation yields which clearly has no solutions (parity incompatibility).
The only cases left to study are (look below)

P.S. : I'm sure there are other less "crude" methods (using divisibility properties of the integers for example)

EDIT :
- For y=1, the only solution is b=5 (and =1) which leads to x=2 or x=3.
- By symmetry, for b=1, the only solution is y=5 which leads us to x=2 or x=3.
- For b=y=2, the only (repeated) solution is x=2
- Take b=2; y=3 we have two solutions x=1 or x=5.
- By symmetry, for b=3; y=2 we also have two solutions x=1 or x=5.
So basically we have 9 solutions : (x,y) { (1;2) ; (1;3) ; (2;1) ; (2;2) ; (2;5) ; (3;1) ; (3;5) ; (5;2) ; (5;3) }

This traduction

[alice02]

Thanks Lostounet!!
I promise you that I will study french!