Question

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

Answer 1

do you mean with integer solutions?

Answer 2

yep.

Answer 3

lets try it

Answer 4

\[17=17\times 1\]
\[(m+n)(m-n)=17\]
\[m+n=17,m-n=1\]
\[m = 9,n=8\]
\[9^2-8^2=17\] works with 17

Answer 5

\[10=5\times 2\]
\[m+n=5,m-n=2\]
oops this won't work because now
\[m=\frac{7}{2}\]

Answer 6

maybe
\[10=10\times 1\]
\[m+n=10,m-n=1\] nope because now
\[m=\frac{11}{2}\]

Answer 7

now it is more or less clear right?

Answer 8

So does it only work with prime numbers?

Answer 9

lets try this one
\[15=5\times 3\]
\[m+n=5,m-n=3\]
\[m=4,n=1\] this one works
\[15=4^2-1^2\]

Answer 10

So is there no general rule to see which numbers this rule works for?

Answer 11

in fact we can do it two ways
\[15=15\times 1\]
\[m+n=15,m-n=1\]
\[m=8,n=7\]
\[15=8^2-7^2\]

Answer 12

no i was just pointing out that it is not a matter of "priime" or not

Answer 13

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
\[m=4\]
\[m=2\]
\[12=4^2-2^2\]