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deena123

  • 3 years ago

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

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  1. deena123
    • 3 years ago
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    @jhonyy9 ??

  2. deena123
    • 3 years ago
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    @PeterPan ??

  3. deena123
    • 3 years ago
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    @Callisto ?? @shadowfiend ??

  4. deena123
    • 3 years ago
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    @hba ??

  5. deena123
    • 3 years ago
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    @Directrix ??

  6. deena123
    • 3 years ago
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    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
    • 3 years ago
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    Pair of zeros satisfies. What next?

  8. deena123
    • 3 years ago
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    ya i know only a pair of zero is the answer but how to get it

  9. deena123
    • 3 years ago
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    i got quadratic residue to be 0,1,2,4

  10. deena123
    • 3 years ago
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    \[x=7x_1,=7x_2\]

  11. deena123
    • 3 years ago
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    @shubhamsrg ?

  12. shubhamsrg
    • 3 years ago
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    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
    • 3 years ago
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    @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
    • 3 years ago
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    hmm, yes you are right. My method is thus flawed.

  15. shubhamsrg
    • 3 years ago
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    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.

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