## goformit100 2 years ago Show that there is no integer a such that a^2−3a−19 is divisible by 289.

1. goformit100

@satellite73

2. AravindG

Are you sure your question right?

3. goformit100

Ya I typed the question Now correctly, Just Accept Answer

4. AravindG

ok How much have you progressed in this question?

5. AravindG

we cant JUST give YOU the answer

6. deathgrowl

did you mean the factor is not integer?

7. goformit100

Sir I am unable to Visulize this Question

8. goformit100

@NotTim

9. goformit100

@Machida

10. NotTim

ah, if only i could help.

11. NotTim

sorry, what's an integer?

12. Machida

|dw:1367902685272:dw|

13. NotTim

uh. how'd you get that?

14. Machida

|dw:1367902803183:dw|

15. goformit100

16. Machida

then a^2-3a+(-19-289k)=0 Do you know that Ax^2+Bx+C=0? I use this B^2 -4AC A=1, B=-3, C=(-19-289k) So its be (-3)^2-4.(1).((-19-289k)

17. goformit100

OK

18. Machida

9+76+1156k 85+1156k then I use 17(5+68k)

19. Machida

5+68k must be have 17 factor for some perfect square, 5+68k=17n 5=17n-68k 5/17=(n-4k) Found that its not integer, because there are 5/17.. even right side is integer. CMIWW @goformit100