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adnanchowdhury
Group Title
Does the difference of two squares identity apply only to algebraic variables? Can we not express any known integer as difference of 2 squares?
 2 years ago
 2 years ago
adnanchowdhury Group Title
Does the difference of two squares identity apply only to algebraic variables? Can we not express any known integer as difference of 2 squares?
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
do you mean with integer solutions?
 2 years ago

adnanchowdhury Group TitleBest ResponseYou've already chosen the best response.0
yep.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
lets try it
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[17=17\times 1\] \[(m+n)(mn)=17\] \[m+n=17,mn=1\] \[m = 9,n=8\] \[9^28^2=17\] works with 17
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[10=5\times 2\] \[m+n=5,mn=2\] oops this won't work because now \[m=\frac{7}{2}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
maybe \[10=10\times 1\] \[m+n=10,mn=1\] nope because now \[m=\frac{11}{2}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
now it is more or less clear right?
 2 years ago

adnanchowdhury Group TitleBest ResponseYou've already chosen the best response.0
So does it only work with prime numbers?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
lets try this one \[15=5\times 3\] \[m+n=5,mn=3\] \[m=4,n=1\] this one works \[15=4^21^2\]
 2 years ago

adnanchowdhury Group TitleBest ResponseYou've already chosen the best response.0
So is there no general rule to see which numbers this rule works for?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
in fact we can do it two ways \[15=15\times 1\] \[m+n=15,mn=1\] \[m=8,n=7\] \[15=8^27^2\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
no i was just pointing out that it is not a matter of "priime" or not
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
lets try 12 \[12=4\times 3\] \[m+n=4,mn=3\] \[2m=7\]no \[12=12\times 1\] \[m+n=12,mn=1\] \[2m=13\] no \[12=6\times 2\] \[m+n=6,mn=2\] \[2m=8\] yes \[m=4\] \[m=2\] \[12=4^22^2\]
 2 years ago
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