Find all integer number (positive) for n, such that :
n^4 - 4n^3 + 22n^2 - 36n + 18 are perfect square numbers!

Question

Find all integer number (positive) for n, such that :
n^4 - 4n^3 + 22n^2 - 36n + 18 are perfect square numbers!

raden · Answer

@mukushla, help me

anonymous · Answer

i will if i can

raden · Answer

i got one of solution is n=1, but any else i dont know....

anonymous · Answer

man wait ... im getting something interesting

anonymous · Answer

$$n^4 - 4n^3 + 22n^2 - 36n + 18=(n^2-2n+9)^2-63$$

anonymous · Answer

$$(n^2−2n+9)^2−63=a^2-63=b^2$$$$a^2-b^2=63$$it is possible only when $a=8$ and  $b=1$

anonymous · Answer

so $n=1$ is the only answer

raden · Answer

hmmm, very interesting ur way for me... because i got my answer with trial and error ^^
but after i see your answer, mukushla i got the answer else :
a^2 - b^2 = 63 (with a=n^2-2n+9)
(a+b)(a-b) = 63
i take 2 case :
=> for the case : (a+b)(a-b) = 9*7
so, a+b=9 and a-b=7 (or reverse)
i got a=8 (got n=1)
=> for the case : (a+b)(a-b) = 3*21
so, a+b=3 and a-b=21 (or reverse)
i got a=12
thus, n^2-2n+9 = 12
n^2-2n-3 = 0
(n-3)(n+1) = 0
n=3 (n=-1 cant)
So, the solutions for n is {1,3)
Isn't right mukushla?

anonymous · Answer

oh yes thats right...silly mistake from my side

raden · Answer

alhamdulillah, thanks for ur help mukushla... 
i cant solve this problem, if i dont see ur step the first... hehe

anonymous · Answer

np man...:)