If $a, b$ are positive integers such that $a^2+2ab-3b^2-41=0$, find the value of $a^2+b^2$
Solution:
The above has $\Delta=4b^2+12b^2+164=4(4b^2+41)$ $\Leftrightarrow$ $4b^2+41=m^2$. For $\color{red}{b\ge11\ \text{it is}\ (2b+1)^2>\Delta>(2b)^2}$, contradiction. So checking the cases we get $b=10$. So the initial equation becomes: $a^2+20a-341=0$, which has the roots: $a=11,31$ . We get the positive one. So, it is $\boxed{a^2+b^2=121+100=221}$}
Find $x$ positive integer such that $x^4+x^3+x^2+x+1$ is a perfect square
Solution:
Note that the problem is equivalent to finding integer solutions to $$4y^2=4x^4+4x^3+4x^2+4x+4$$ Now proceed to note that if $\color{red}{x>3,\text{we can find}}$ $$\color{red}{(2x^2+x)^2=4x^4+4x^3+x^2 < 4x^4+4x^3+4x^2+4x+4=4y^2}$$ And $$4y^2 < 4x^4+4x^3+5x^2+2x+1=(2x^2+x+1)^2 $$ Since $4x^4+4x^3+4x^2+4x+4$ is stuck between squares of two consecutive numbers, it cannot be a square itself, which is a contradiction.
Thus we have that if $x$ is a positive integer it must be less than, or equal to, $3$. Trial and error gives us $x=3, y=11 $ is a valid solution.
I got 1st from Aops and 2nd from MSE.
My problem is how do they get those bounds ($x>3,b\ge11$), what is the Algorithm, and how to verify that bound is correct, please help.
(see the highlighted part)
