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

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deena123 Group TitleBest ResponseYou've already chosen the best response.0
@jhonyy9 ??
 one year ago

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@PeterPan ??
 one year ago

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@Callisto ?? @shadowfiend ??
 one year ago

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@Directrix ??
 one year ago

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

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Pair of zeros satisfies. What next?
 one year ago

deena123 Group TitleBest ResponseYou've already chosen the best response.0
ya i know only a pair of zero is the answer but how to get it
 one year ago

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i got quadratic residue to be 0,1,2,4
 one year ago

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\[x=7x_1,=7x_2\]
 one year ago

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@shubhamsrg ?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
@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.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
hmm, yes you are right. My method is thus flawed.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
how about if we represent it like this: 7(a^2+ b^2) = (m+ni)(mni) 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)(piq) a^2 + b^2 = (a+bi) (abi) now this process goes on forever.
 one year ago
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