Hi everyone,
I am certain you have come across this before. It is that you can express odd and even (that have a factor of 4) numbers as the difference of two positive integer squares.
I can prove for both cases that indeed the proposition is valid. However I cannot show why even numbers that do not divide fully into 4 cannot be expressed as the difference of two positive integer squares.
Can someone help me out? Google is not of much help.
Here is what I have noticed:
Now if n has a factor of 4, then (a-b) is even and (a+b) is even too. For odd numbers, (a-b) always = 1, and therefore (a+b) = odd (i.e. 7=(4-3)(4+3)=1*(4+3)=1*7=7).
However numbers like 2,6,10 (cannot be expressed as the difference of two positive integer squares) are derived by multiplying an odd and an even, e.g. 2=2*1, 6=2*3, 10=2*5;
I wonder whether I just show that since (a+b)>(a-b) then (a+b) = 2 and (a-b) = 1 for n = 2, and similarly (a+b)=3 and (a-b) = 2 for n = 6 etc. That there are no integer solutions for a and b? How would I generalize this result?
P.S. Bonus question How can I show that \(\displaystyle (a+b) \geq \sqrt{n}\) and hence \(\displaystyle (a-b) \leq \sqrt{n} \)
Thank you
I am certain you have come across this before. It is that you can express odd and even (that have a factor of 4) numbers as the difference of two positive integer squares.
I can prove for both cases that indeed the proposition is valid. However I cannot show why even numbers that do not divide fully into 4 cannot be expressed as the difference of two positive integer squares.
Can someone help me out? Google is not of much help.
Here is what I have noticed:
\(\displaystyle a^2-b^2=(a-b)(a+b)=n\)
Now if n has a factor of 4, then (a-b) is even and (a+b) is even too. For odd numbers, (a-b) always = 1, and therefore (a+b) = odd (i.e. 7=(4-3)(4+3)=1*(4+3)=1*7=7).
However numbers like 2,6,10 (cannot be expressed as the difference of two positive integer squares) are derived by multiplying an odd and an even, e.g. 2=2*1, 6=2*3, 10=2*5;
I wonder whether I just show that since (a+b)>(a-b) then (a+b) = 2 and (a-b) = 1 for n = 2, and similarly (a+b)=3 and (a-b) = 2 for n = 6 etc. That there are no integer solutions for a and b? How would I generalize this result?
P.S. Bonus question How can I show that \(\displaystyle (a+b) \geq \sqrt{n}\) and hence \(\displaystyle (a-b) \leq \sqrt{n} \)
Thank you