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

1. sauravshakya

One is i=2 and j=5

2. sauravshakya

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

4. sauravshakya

Looks like 4 solution.

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

6. sauravshakya

|dw:1350668697206:dw|

oh thanks i see my folly i did not factor 2

8. sauravshakya
9. sauravshakya

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

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

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

yes

13. Zekarias

ok

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

15. Zekarias