Find all pears of (a,b) satisfy :
a^4 - 49a^2 + 1 = b^2. (a,b are natural number)

Question

Find all pears of (a,b) satisfy :
a^4 - 49a^2 + 1 = b^2. (a,b are natural number)

anonymous · Answer

i think one of solution is (7,1), but i got the answer using trial and error method :)
what about u @mukushla ??

anonymous · Answer

im workin on it...

one method is making a quadratic equation in terms of  $a^2$ ...then the Discriminant of that quadratic must be perfect square...let me check it

anonymous · Answer

let  $a^2=m$  then u have$$m^2 - 49m + 1 - b^2=0$$Discriminant= $49^2-4(1-b^2)=n^2$ must be a perfect square it gives$$(n-2b)(n+2b)=2397=3*17*47$$ now there are few options :

1............$n-2b=1$  and  $n+2b=2397$
2............$n-2b=3$  and  $n+2b=799$
3............$n-2b=17$  and  $n+2b=141$
4............$n-2b=47$  and  $n+2b=51$

anonymous · Answer

Why discriminant of that quadratic must be perfect square??

anonymous · Answer

ohh i see, so that the roots of quadtic equation be integer, isn't ?

anonymous · Answer

yes http://openstudy.com/study#/updates/50028852e4b0848ddd669ca2

anonymous · Answer

lets solve it completely...

anonymous · Answer

i late get information from u mukushla... :D
So, if i solve the last that question. 
I got (n,b) = (1199,599), (401, 199), (79,31), and (49,1)
what next i do??

anonymous · Answer

now we must plug values of $b$ in the quadratic and solve for $m$ .... $m$ must be a perfect square too because we solved equation with  $m=a^2$

anonymous · Answer

solve and tell me what u get?

anonymous · Answer

only m?

anonymous · Answer

$b$ gives  $m$ and  $m$  gives  $a$

anonymous · Answer

not conencted in my mind... (b gives  m and  m  gives  a, what mean is it)

anonymous · Answer

subtitute for b to m^2−49m+1−b^2=0 ?

anonymous · Answer

yeah

anonymous · Answer

if b=1, i got m^2-49m=0
=> m(m-49)=0
m=0 (not satisfy)
m=49
from a^2 = m
a^2 = 49, so a=7 (-7 not natural number)
so the first pear for (a,b) = (7,1)
and next next next .... right mukushla?

anonymous · Answer

yep....:D

anonymous · Answer

i was just solved it, hehe... thanks million for u mukushla...

anonymous · Answer

You are Welcome

anonymous · Answer

but im not sure i can solve again if given like this problems... :D

anonymous · Answer

im sure u can...

anonymous · Answer

if the problems same like this, hehe
but i will study hard to get it......