$(4k-1)^2 +(4k)^2$ is a perfect square Let $k$ be strictly bigger than $1$. 
Is there any integer $k$ such that $(4k-1)^2+(4k)^2$ is a perfect square?
My computation shows that there are infinitely many such $k$, namely those arising from the Pell
equation $X^2-2Y^2=-1$.
However the solution manual says that there are no such $k$.
Could anyone tell me the right answer and solution preferrably without using Pell equation?
 A: If you have a solution $(k,w)$ to $$ (4k-1)^2 + (4k)^2 = w^2,  $$
you get the next larger $(k,w)$ pair with
$$  (17 k + 3 w - 2, 96 k + 17 w - 12)   $$
So the $(k,w)$ pairs are
$$ (1,5),  $$
$$ (30,169),  $$
$$ (1015,5741),  $$
$$ (34476,195025),  $$
$$ (1171165,6625109),  $$
$$ (39785130,225058681),  $$
$$ (1351523251,7645370045),  $$
and so on.
There are no Pell equations here. These are not the droids you seek. Attention Deficit Ooh Shiny!
A: For $k=30$, you have $119^2+120^2=169^2$
A: There is indeed a Pell equation involved here.  When two adjacent numbers whose square sums to a square, the octagonal series is involved.  This approximates $\sqrt{2}$, for which the relevant number is 6.
  0    1   2    5   12   29   70   169    x   X=x+y
  1    1   3    7   17   41   99   239    y   Y=x+X

The solutions you seek are to be found when the first number is an odd number in the top row.  The case here is $2*5*12=120$ and $7*17=119$, these add up to some odd number in the top row, here $169^2$.
The series might be found, by $c(n+1)= 6 \cdot c(n) - c(n-1)$,  the series here is $(A*B)^2 + (2*C*D)^2$  
   A   1    7   41      239      1393
   B   3   17   99      577      3363
   C   1    5   29      169       985
   D   2   12   70      408      2378 
   E   5  169  5741  195025   6625109

The final square can be derived by a similar series to above, by replacing $6$ by $34$, which gives the values in E in the table above.  
