## deena123 2 years ago How many pairs of integers are there such that \[a^2+6b^2\] and \[b^2+6a^2\] are perfect squares?

1. deena123

@jhonyy9 ??

2. deena123

@PeterPan ??

3. deena123

4. deena123

@hba ??

5. deena123

@Directrix ??

6. deena123

i took \[a^2+6b^2=x^2 \]\[b^2+6a^2=y^2\]\[7(a^2+b^2)=x^2+y^2\]

7. klimenkov

Pair of zeros satisfies. What next?

8. deena123

ya i know only a pair of zero is the answer but how to get it

9. deena123

i got quadratic residue to be 0,1,2,4

10. deena123

\[x=7x_1,=7x_2\]

11. deena123

@shubhamsrg ?

12. shubhamsrg

let a>=b ,and a^2 + 6b^2 = m^2 b^2 + 6a^2 = n^2 , for some integers m and n. also, =>7(b^2 + a^2) = m^2 + n^2 -->b^2 + a^2 = (m^2 /7) + (n^2 /7) thus both m^2 and n^2 must be divisible by 7 for LHS to be an integer. let m=7p and n=7q => b^2 + a^2 = 7(p^2 + q^2) again the same thing i.e. b^2 and a^2 should be divisible by 7. this process will go on forever, I am unable to highlight the conclusion as to why (0,0) is the only solution, though logically, its quite evident that it is the only soln.

13. klimenkov

@shubhamsrg That is not right. If the sum of two numbers is divisible by 7 that is not right that any of this numbers must be divisible by 7: 7b=20+1 b=3, but we could not say that 20 is divisible by 7.

14. shubhamsrg

hmm, yes you are right. My method is thus flawed.

15. shubhamsrg

how about if we represent it like this: 7(a^2+ b^2) = (m+ni)(m-ni) 7 is a prime number. hence 7 should divide both m and n in this case. m=7p and n=7q (a^2+ b^2) = 7(p+iq)(p-iq) a^2 + b^2 = (a+bi) (a-bi) now this process goes on forever.