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how many primes are there in the form\[m^4+4n^4\]
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Let p =m^4 + 4n^4 THEN p=(m^2+2n^2-2mn)(m^2+2n^2+2mn)
where P is a prime number....... NOW, ((m^2+2n^2-2mn)=1 AND (m^2+2n^2+2mn)=p
Am i doing correct?
yup
and then?
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So, we need to solve that two equations
Further simplification will get 1+4mn=p
So, m=1 and n=1 is one of the solution
emm...dont look for p prime number cant be has 2 factors so just look for m,n from m^2+2n^2-2mn=1
m^2+2n^2-2mn=1 (m^2-2mn+n^2)+n^2=1 (m-n)^2=1-n^2 m=n+- (1-n^2)^1/2
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So, n cannot be greater than 1
Thus, the only solution is when n=1
n=m=1 and p=5 well done saura
Thanx
note that when one of m or n is greater than 1 then m^2+2n^2-2mn>1
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so both of them must be 1
Oh ya.... that way can also be done
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