Fermat showed that no two fourth powers can ever sum to a perfect square.

Go one better by proving that there is no solution in the positive integers for the equation a^4+b^4+1 = c^2.

Or give a counter-example.

(Acknowledgement: suggested by a post of Daniel's on one of my earlier problems)

The approach is to consider the case where the 4th powers differ by one, then extend this to larger differences. We are going to show that if we do this, the so-called 'x^2' term is one more than a 4th power, rather than the desired square.

I  Note that b^4+(b+1)^4+1 = 2(b^2+b+1)^2, say, 2P^2. We can increase the difference between the 4th powers by adding a variable c to the second one: (b+c+1)^4.

II (b^4+(b+c+1)^4+1) - (b^4+(b+1)^4+1) = (b+c+1)^4-(b+1)^4 (simple cancelling); 

hence (b^4+(b+c+1)^4+1) - 2P^2 = (b+c+1)^4-(b+1)^4, say, Q^4-R^4

III    So
 (b^4+(b+c+1)^4+1) = (b+c+1)^4-(b+1)^4 +2(b^2+b+1)^2
 (b^4+(b+c+1)^4+1) = Q^4-R^4+2P^2

IV  Re-substituting in x^2 =  Q^4-R^4+2P^2:

x^2+(b+1)^4 = (b+c+1)^4 + 2(b^2+b+1)^2

V  We can do something with this, because if we set x^2=(y^4+1), then there are at once integer solutions for {b= (-y) , c= (y-1)}, { b=0, c=(-y-1)}, { b=0, c=(y-1)}, {b=(y+1), c=(-y-1))}, y in N. (Obviously, there are no solutions with both b,c, positive - Fermat):

V(i) (y^4+1)+(-y+1)^4 =  (-y+(y-1)+1)^4 + 2((-y)^2+-y+1)^2 (compare I above)
V(ii) y^4+1+(0+1)^4    =   (0+(-y-1)+1)^4 + 2(0^2+0+1)^2    
V(iii) y^4+1+(0+1)^4    =   (0+(y-1)+1)^4 + 2(0^2+0+1)^2
V(iv) y^4+1+(y+1)^4 = (y+(-y-1)+1)^4 + 2(y^2+y+1)^2 (compare I above)

VI The snag is that there are no non trival 4th powers that differ from a square by 1. If we seek an equality, we must settle for at least twice a square, as in I, V(i), and V(iv) above.

Edited on December 27, 2012, 12:03 am

Posted by broll on 2012-12-26 02:52:01