Anyone can give me hints on these...

M

Mono Yung

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I am not really know the solutions, can anyone give me some hints to solve these problems?
Many thanks!
 
I really like that 1st problem and since it has been more than 3 1/2 days since we heard from Mono I will post my solution.

a^2 + 2b^2 = c^2 <=> 2b^2 = c^2 - a^2 = (c+a)(c-a). Since the lhs is even it is clear that the rhs is also even.

So c+a or c-a must be even. If c+a is even then c and a both have the same parity. Similarly if c-a is even then c and a have the same parity. TSince c and a have the same parity it follows that c+a and c-a are even. The product of two even numbers is a multiply of 4. So 2b^2 is a multiply of 4 => b^2 is a multiply of 2=>b is a multiply of 2 => b is even
 
Somehow (probably because the baby needed my attention) I failed to post that there is an error in problem 6.

[MATH]a_1 = 2.[/MATH]
I think there are probably two ways to prove this, but I suspect induction may be the easier method.
 
Jomo said:
The product of two even numbers is a multiply of 4. So 2b^2 is a multiply of 4 => b^2 is a multiply of 2=>b is a multiply of 2 => b is even
Click to expand...


Jomo, the word that is needed here four times is spelled as "multiple."
 
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