Here's the question you clicked on:

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

• This Question is Closed
1. satellite73

do you mean with integer solutions?

2. adnanchowdhury

yep.

3. satellite73

lets try it

4. satellite73

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

5. satellite73

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

6. satellite73

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

7. satellite73

now it is more or less clear right?

8. adnanchowdhury

So does it only work with prime numbers?

9. satellite73

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

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

11. satellite73

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. satellite73

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

13. satellite73

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 $m=4$ $m=2$ $12=4^2-2^2$