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the question is a little unclear, but maybe it is this: all triples can be found by taking whole number \(m,n\) and computing \[m^2-n^2, 2mn, m^2+n^2\] is that what you have to find, \(m\) and \(\)?
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Thank you cx
its apex so the answers would be like A.x=4,y=3 B.X=1, y+3 C.x=3,y=2 D.x=2 y=2
what does "it is apex" mean?
and thanks for the warm welcome and its an online class
ok it looks like they used \(x\) and \(y\) whereas i used \(m\) and \(n\) but it is the same thing
what would be the answer?
you have \[x^2-y^2=7\] that is the same as \[(x-y)(x+y)=7\] there is only one way to factor \(7\) as \(1\times 7\) so \[x-y=1\\ x+y=7\] and using that you can solve for \(x\) and \(y\)
so i basically plug in and solve
you can solve that in your head: two numbers that add to 7 and are one apart
ooh i see A is \(x=4,y=3\) yes, that is right