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Mathematics 20 Online
OpenStudy (anonymous):

how many primes are there in the form\[m^4+4n^4\]

OpenStudy (anonymous):

Let p =m^4 + 4n^4 THEN p=(m^2+2n^2-2mn)(m^2+2n^2+2mn)

OpenStudy (anonymous):

where P is a prime number....... NOW, ((m^2+2n^2-2mn)=1 AND (m^2+2n^2+2mn)=p

OpenStudy (anonymous):

Am i doing correct?

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

and then?

OpenStudy (anonymous):

So, we need to solve that two equations

OpenStudy (anonymous):

Further simplification will get 1+4mn=p

OpenStudy (anonymous):

So, m=1 and n=1 is one of the solution

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

So, n cannot be greater than 1

OpenStudy (anonymous):

Thus, the only solution is when n=1

OpenStudy (anonymous):

n=m=1 and p=5 well done saura

OpenStudy (anonymous):

Thanx

OpenStudy (anonymous):

note that when one of m or n is greater than 1 then m^2+2n^2-2mn>1

OpenStudy (anonymous):

so both of them must be 1

OpenStudy (anonymous):

Oh ya.... that way can also be done

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