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 one year ago
How many pairs of integers are there such that \[a^2+6b^2\] and \[b^2+6a^2\] are perfect squares?
 one year 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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deena123
 one year ago
Best ResponseYou've already chosen the best response.0@Callisto ?? @shadowfiend ??

deena123
 one year ago
Best ResponseYou've already chosen the best response.0i took \[a^2+6b^2=x^2 \]\[b^2+6a^2=y^2\]\[7(a^2+b^2)=x^2+y^2\]

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

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

deena123
 one year ago
Best ResponseYou've already chosen the best response.0i got quadratic residue to be 0,1,2,4

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2let 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.

klimenkov
 one year ago
Best 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.

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

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.2how 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.
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