Fourth Power Differences

Note that the difference between consecutive 4th powers increases as the consecutive numbers increase. This is obvious.

We see that a solution to $(x,y)=(2,1)$ . Note that this means that there aren't any bigger solutions, because the difference between consecutive 4th powers must increase.

Are we done? No, because $x,y$ can also be negative. Thus our solutions are actually $(x,y)=(\pm2,\pm 1)$ so there are $\boxed{4}$ solutions.

Note that $x^{4}-y^{4}=(x^{2}-y^{2})(x^{2}+y^{2})$ upon factorisation. Since $x^{2}>y^{2}>0$ ,

$0<x^{2}-y^{2}<x^{2}+y^{2}$ .

Clearly, both $x^{2}-y^{2}$ and $x^{2}+y^{2}$ are positive integers. Since $15=1\times15=3\times5$ , we find that only $x^{2}-y^{2}=3$ and $x^{2}+y^{2}=5$ yields $y=\pm1$ and $x=\pm2$ (In the other case, both $x$ and $y$ are not integers). Hence, the total number of integer solutions is $2\times2=\boxed{4}$ solutions.

Write in a form of completing the square:

$x^4-y^4-15=0$

Now, from here, we may split the 15 to either x or y as a perfect integer 4th. Perfect integer 4ths up to 15 are:

$1^4=1$ , $2^4=16$ , etc. Notice that we can move past 15, since via algebraic manipulation we can move things around a bit. Ex.:

$(x^4-16)-(y^4-1)=0$ ,

Also note, that you cannot swap integer partners for x and y, because that will yield an incompatible combination ( $(x^4-1)-(y^4+???)=0$ ,

Notice that possibilities such as

$(x^4-1)-(y^4+14)=0$ ,

don't count, because even though the x part is satisfied, y is left in soil.

Trying further possible arrangements will show that the 4th deg. differences grow too great:

$(x^4-81)-(y^4+???)=0$ ...

Note that the constant terms must add to 15. Trying 4th degree integers greater than 2 will yield a difference too great, and thus, leaving us with only 4 possibilities:

$x=-2, 2; y=-1, 1$

The answer is $\boxed{4}$

ngomong opo kuwi. dellok wolfram wae to dab

the largest degree represents the number of solutions.