anonymous
  • anonymous
Does the difference of two squares identity apply only to algebraic variables? Can we not express any known integer as difference of 2 squares?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
do you mean with integer solutions?
anonymous
  • anonymous
yep.
anonymous
  • anonymous
lets try it

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anonymous
  • anonymous
\[17=17\times 1\] \[(m+n)(m-n)=17\] \[m+n=17,m-n=1\] \[m = 9,n=8\] \[9^2-8^2=17\] works with 17
anonymous
  • anonymous
\[10=5\times 2\] \[m+n=5,m-n=2\] oops this won't work because now \[m=\frac{7}{2}\]
anonymous
  • anonymous
maybe \[10=10\times 1\] \[m+n=10,m-n=1\] nope because now \[m=\frac{11}{2}\]
anonymous
  • anonymous
now it is more or less clear right?
anonymous
  • anonymous
So does it only work with prime numbers?
anonymous
  • anonymous
lets try this one \[15=5\times 3\] \[m+n=5,m-n=3\] \[m=4,n=1\] this one works \[15=4^2-1^2\]
anonymous
  • anonymous
So is there no general rule to see which numbers this rule works for?
anonymous
  • anonymous
in fact we can do it two ways \[15=15\times 1\] \[m+n=15,m-n=1\] \[m=8,n=7\] \[15=8^2-7^2\]
anonymous
  • anonymous
no i was just pointing out that it is not a matter of "priime" or not
anonymous
  • anonymous
lets try 12 \[12=4\times 3\] \[m+n=4,m-n=3\] \[2m=7\]no \[12=12\times 1\] \[m+n=12,m-n=1\] \[2m=13\] no \[12=6\times 2\] \[m+n=6,m-n=2\] \[2m=8\] yes \[m=4\] \[m=2\] \[12=4^2-2^2\]

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