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adnanchowdhury

  • 4 years ago

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

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  1. anonymous
    • 4 years ago
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    do you mean with integer solutions?

  2. adnanchowdhury
    • 4 years ago
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    yep.

  3. anonymous
    • 4 years ago
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    lets try it

  4. anonymous
    • 4 years ago
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    \[17=17\times 1\] \[(m+n)(m-n)=17\] \[m+n=17,m-n=1\] \[m = 9,n=8\] \[9^2-8^2=17\] works with 17

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

  6. anonymous
    • 4 years ago
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    maybe \[10=10\times 1\] \[m+n=10,m-n=1\] nope because now \[m=\frac{11}{2}\]

  7. anonymous
    • 4 years ago
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    now it is more or less clear right?

  8. adnanchowdhury
    • 4 years ago
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    So does it only work with prime numbers?

  9. anonymous
    • 4 years ago
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    lets try this one \[15=5\times 3\] \[m+n=5,m-n=3\] \[m=4,n=1\] this one works \[15=4^2-1^2\]

  10. adnanchowdhury
    • 4 years ago
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    So is there no general rule to see which numbers this rule works for?

  11. anonymous
    • 4 years ago
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    in fact we can do it two ways \[15=15\times 1\] \[m+n=15,m-n=1\] \[m=8,n=7\] \[15=8^2-7^2\]

  12. anonymous
    • 4 years ago
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    no i was just pointing out that it is not a matter of "priime" or not

  13. anonymous
    • 4 years ago
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    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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