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this one is about factoring also...but a little bit nicer and harder\[p^x=y^4+4\]\(p\) is a prime...\(x,y\) are natural.
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What are we supposed to do?
well...find all solutions
this one is not mine
it will be always odd because lhs is an odd no
Oh.... got it can be both
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how...lhs is (prime no)^natural no....so odd ...right?
p=2
oh...yes...for p=2 it'll fail
1...factor y^4+4 2...p^x=ab then a,b are some exponents of p
y^4+4=(y^2-2y+2)(y^2+2y+2)
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