X = n6 + m11

(n , m are non-negative integers)

Let N be greater than or equal to 50 and satisfy:

N = u6 + v11

(u , v are non-negative integers)

we know something more about u and v, because N>= 50

impossible:

u = 0, v =0,1,2,3,4

u = 1, v =0,1,2,3

u = 2, v =0,1,2,3

u = 3, v= 0,1,2

u = 4, v= 0,1,2

u = 5, v= 0,1

u = 6, v= 0,1

u = 7, v= 0

u = 8, v= 0

we now have to show that N+1 one can be writen in this form:

N+1 = u6 + v11 +1

say u =0 , and v >= 5 then v = 5 + z (z=0,1,2,..)

N+1 = v11 +1 = (5+z)11+1 = (4+z)11 + 12

= 2.6 + (4+z)11

say u =1, and v >=4 then v = 4+ z (z=0,1,2,..)

N+1 = 6 + (4+z) 11 +1 = 6 + (3+z) 11 + 12

= 3.6 + (3+z) 11

etc, etc

u = 2: 12 + (4+z) 11 +1 = 4.6 + (3+z) 11

u = 3: 18 + (3+z) 11 + 1 = 5.6 + (2+z) 11

u = 4: 24 + (3+z) 11 +1 = 6.6 + (2+z) 11

u = 5: 30 + (2+z) 11 +1 = 7.6 + (1+z) 11

u = 6: 36 + (2+z) 11 +1 = 8.6 + (1+z) 11

u = 7: 42 + (1+z) 11 +1 = 9.6 + z 11

u = 8: 48 + (1+z) 11 +1 = 10.6 + z 11

(The same now has to be shown with v =0,1,2,3,..)

now find u,v , such that:

50 = u6 + v11

one solution: u=1, v = 44

—————-

(I’m not really sure about all of this, good luck)

It may be beter and easier to write it in a more general way

say N satisfies:

N = u 6 + v 11

then

N+1 = u 6 + v 11 + 1 = u 6 + (v-1) 11 + 12 = (u +2) 6 + (v-1) 11