[Guest]

Determine all possible pair(s) (X, Y) of nonnegative integers that satisfy this equation.

X² + (X+1)² = Y⁴ + (Y+1)⁴

[Guest]

the trivial solution is x,y = 0,0. I'm not positive, but this may be the only solution.

[superprismatic]

Since the function F(X,Y)=Y⁴+(Y+1)⁴-X²-(X+1)²

is a differentiable function of 2 variables, an extremum point can be found in the usual

way by taking partials, setting them to 0, and solving. One must also check boundary

points which in this case is (X,Y)=(0,0). Doing this, we find that F(X,Y) is minimal

at only the boundary point. There are other candidates from doing the partial derivatives,

but they are not non-negative integers. So, (0,0) is the only solution.

[Guest]

i checked up to 100,000, as far as i can tell there is no other solution.

[superprismatic]

Since the function F(X,Y)=Y⁴+(Y+1)⁴-X²-(X+1)²

is a differentiable function of 2 variables, an extremum point can be found in the usual

way by taking partials, setting them to 0, and solving. One must also check boundary

points which in this case is (X,Y)=(0,0). Doing this, we find that F(X,Y) is minimal

at only the boundary point. There are other candidates from doing the partial derivatives,

but they are not non-negative integers. So, (0,0) is the only solution.

Well, I forgot to actually check the extremum candidate from the partials. It would have

to make F(X,Y) greater than 0 for my remarks to make sense, i.e. for the solution over

the reals to be no better than the boundary candidate. As it is, the candidate from the

partials makes F(X,Y) take its smallest value of -3/8 at (-1/2,-1/2). So, my previous

post is incorrect.

[superprismatic]

i checked up to 100,000, as far as i can tell there is no other solution.

Did you check to see what's the closest you got to a solution, i.e., what is the closest the right and left hand sides got to each other? It would be interesting to know! If you got as close as 1 or 2 or so, you may not have gone out far enough.

[Guest]

just checked and the greater the y the less close x can get as far as i can tell.

[Guest]

I too think (0,0) is the only solution.

The given expression can be re-written as:

(X - Y²).(X + Y²) = - ((X - Y²) - 2.Y).((X + Y²) + 2Y + 2)

Since signs on both sides need to be the same, and since X and Y themselves are positive, we have two conditions (ignoring the said solution where both sides are zero):

Both sides are positive sign:

=> X > Y² (and, also, X < Y² + 2.Y, but that doesn't matter)

Both sides are negative sign:

=> X < Y² (to make L.H.S negative). But this would imply that the R.H.S will become positive. So this is an impossible condition.

But if, X > Y², the original equality can't be satisfied.

So no other solutions are possible

[Guest]

Determine all possible pair(s) (X, Y) of nonnegative integers that satisfy this equation.

X² + (X+1)² = Y⁴ + (Y+1)⁴

X² + (X+1)² = Y⁴ + (Y+1)⁴

2X² + 2X + 1 = 2Y⁴ + 4Y³ + 6Y² + 4Y +1

2X² + 2X + 1 - 1 = 2Y⁴ + 4Y³ + 6Y² + 4Y

2X (X +1) = 2Y (Y³ + 2Y² + 3Y + 2)

2X (X +1) = 2Y (Y+1)(Y² + Y + 2)

so your only solutions for X are {0,-1} and Y {0,-1, and some rational}

the only non-neg integer solution for (X,Y) is (0,0)

[Guest]

I'm sorry but I think 'methinks' and 'Perigren' made mistakes in their proofs.

Ok here is why and my 2 cents:

first i agree with you 'methinks' up until you say "But if, X > Y², the original equality can't be satisfied." because the way I see it, X must be > Y². basically expand the right side of the equation and you have:

2X² + 2X + 1 = 2Y⁴ + 4Y³ + 6Y² + 4Y +1 and now sub Y² in for X and you have:

2Y⁴ + 2Y² + 1 = 2Y⁴ + 4Y³ + 6Y² + 4Y +1 and you can plainly see that the left side is too small, so X must be > Y²

next, I like what you did 'Perigren' up until the end and then it looks like you solved each side being=0, but there is no reason to think that.

however, if I can start at your end point:

2X (X +1) = 2Y (Y+1)(Y² + Y + 2)

and rearrange the right hand side:

X (X +1) = Y(Y+1) (Y(Y + 1) + 2)

now define a = Y(Y+1) and you have:

X (X +1) = a (a+2)

but there is no (positive) integers solution for X and a

ahhhh, now I feel better