Answer
The equation can be written as $x^{4}$ = $5^{4}$ - $y^{4}$.
Therefore $x^{4}$ = (25+$y^{2}$)(25-$y^{2}$)
Now y ranges from 1 to 4 as x,y>0 .
For y=1 we have $x^{4}$=(26)(25). Therefore x is not an integer.
Next for y=2 we have $x^{4}$=(29)(21). Again x is not an integer.
Next for y=3 we have $x^{4}$=(34)(16). Again x is not an integer.
Next for y=4 we have $x^{4}$=(41)(9). Again x is not an integer.
Therefore no integral pair of x and y exists.
Hence no solution to the given equation exists.
Work Step by Step
The equation can be written as $x^{4}$ = $5^{4}$ - $y^{4}$.
Therefore $x^{4}$ = (25+$y^{2}$)(25-$y^{2}$)
Now y ranges from 1 to 4 as x,y>0 .
For y=1 we have $x^{4}$=(26)(25). Therefore x is not an integer.
Next for y=2 we have $x^{4}$=(29)(21). Again x is not an integer.
Next for y=3 we have $x^{4}$=(34)(16). Again x is not an integer.
Next for y=4 we have $x^{4}$=(41)(9). Again x is not an integer.
Therefore no integral pair of x and y exists.
Hence no solution to the given equation exists.
