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Zawadx wrote:Couldn't submit answers due to power failure :/ Can anyone give any tips on how to type out solutions faster?

btw, Vieta Jumping can direct you to the solution for number 4 but I didn't have to use it directly in the solution.

তোমাদের এটা একটা সমস্যা যে তোমরা ল্যাটেক ভাল পারো না, কিন্তু পারা উচিত। পরের কথা হলো তোমাদের সমাধান ও অনেক সময় ফর্মাল হয় না। সেজন্য ফোরামে অনেক সমস্যা আর সমাধান পোস্ট করলেই আস্তে আস্তে প্র্যাকটিস হয়। এটাও ফোরাম ব্যবহারের একটা উদ্দেশ্য। কিন্তু এতদিন তোমরা কেউ কথা কানে নেও নাই। এখন তার জন্য কিছু তো ভোগ করা লাগবেই। আর কোন টিপস মনে হয়না আছে। এখনো শুরু কর, দেরী হয় নাই।

Can p4 be solved like this ? (a^2 - 1)*(b^2-1) = a^2*b^2-a^2-b^2+1 . It is divisible by both a and b . And , a^2*b^2 is divisible by ab . So that leaves 1-a^2-b^2 . Let's multiply it with (-1) . Result = a^2+b^2+1 which is divisible by ab . And as for k being a positive integer , a and b are both greater than 1 . So a^2 and b^2 must be greater than 1 . So a^2+b^2-1 is a positive integer , so is ab . In this way k is a positive integer greater than 1 . Can anyone tell me if it is correct ?

Epshita32 wrote:Can p4 be solved like this ? (a^2 - 1)*(b^2-1) = a^2*b^2-a^2-b^2+1 . It is divisible by both a and b . And , a^2*b^2 is divisible by ab . So that leaves 1-a^2-b^2 . Let's multiply it with (-1) . Result = a^2+b^2+1 which is divisible by ab . And as for k being a positive integer , a and b are both greater than 1 . So a^2 and b^2 must be greater than 1 . So a^2+b^2-1 is a positive integer , so is ab . In this way k is a positive integer greater than 1 . Can anyone tell me if it is correct ?

You are supposed to prove that for _any_ $k>1$, you can find $a$ and $b$ satisfying the equation. You only proved that there are some $k$ for which solutions to the equation exist.

Epshita32 wrote:Can p4 be solved like this ? (a^2 - 1)*(b^2-1) = a^2*b^2-a^2-b^2+1 . It is divisible by both a and b . And , a^2*b^2 is divisible by ab . So that leaves 1-a^2-b^2 . Let's multiply it with (-1) . Result = a^2+b^2+1 which is divisible by ab . And as for k being a positive integer , a and b are both greater than 1 . So a^2 and b^2 must be greater than 1 . So a^2+b^2-1 is a positive integer , so is ab . In this way k is a positive integer greater than 1 . Can anyone tell me if it is correct ?

It's not totally correct but good work! You don't have to get it right at the first attempt. Learn by making mistakes
Here are some extended hints: First understand thoroughly what the problem requires. $\dfrac{a^2+b^2-1}{ab}=k$ must have a solution $(a,b)$ for all $k>1$. You can learn Vietta jumping later. Here is a motivation for the solution without Vietta jumping. Notice that, all you are required to do is finding appropriate representation for $a$ and $b$ so that the denominator and numerator cancels out in way which leaves only $k$. And from $b|a^2-1=(a+1)(a-1)$, it should strike you to take something like $b=a+1$ or $b=a-1$ to see if it works. If it doesn't play around a while with them. Another hint: sometimes you have to raise the powers of variables too.

Epsita, before you solve this problem, try to solve and understand the following problems(these are problems of representation type):
1. Any odd integer can be written as $a^2-b^2$. Just take $a=b+1$ and see that you are done.
2. Any perfect square can be written as $\dfrac{a^3+b^3}{a+b^2}$. If you put some known expressions randomly, you can be lucky since this is an easy one. But here is the intuition: $a+b^2|a^3+b^3$ and $a+b^2|ab+b^3$, so $a+b^2|ab-a^3=a(b-a^2)$. Now take $\gcd(a,b)=1$ so that $\gcd(a+b^2,a)=1$. Then $a+b^2|b-a^2$. Now this tells us to take $b=a^2$ and indeed it works for all perfect squres(check it!). Hope these help