## anonymous 3 years ago j^3 = 61 + i^6 Find all the integer solution of i and j.

1. anonymous

One is i=2 and j=5

2. anonymous

3. anonymous

$j^3-(i^2)^3=61$ $(j-i^2)(j^2+ij+i^2)=61$ 61 is prime so 61(1) are the only factors hence $j-i^2=1,j^2+ij+i^2=61$

4. anonymous

Looks like 4 solution.

5. anonymous

but we imgore the fact that it can be vise versa 1(61)

6. anonymous

|dw:1350668697206:dw|

7. anonymous

oh thanks i see my folly i did not factor 2

8. anonymous
9. anonymous

Looks like 2 and 5 only.......... But how to prove it mathematically.

10. anonymous

$i^4+2i^2+1+i^2(1+i^2)+i^4=61$ $3i^4+3i^2-60=0,i^4+i^2-10=0$

11. anonymous

sorry $i^4+i^2-20$

12. anonymous

yes

13. anonymous

ok

14. anonymous

$i=\pm 2,j=(\pm2)^2+1=5$

15. anonymous