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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?

Mathematics
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Other answers:

@hba ??
i took \[a^2+6b^2=x^2 \]\[b^2+6a^2=y^2\]\[7(a^2+b^2)=x^2+y^2\]
Pair of zeros satisfies. What next?
ya i know only a pair of zero is the answer but how to get it
i got quadratic residue to be 0,1,2,4
\[x=7x_1,=7x_2\]
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.
@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.
hmm, yes you are right. My method is thus flawed.
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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