$x^2+x+100 = k^2$, $(x, k) \in \mathbb{N}^2$

Question

How can I solve this diophantine equation ? $$x^2+x+100 = k^2 ,(x, k) \in \mathbb{N}^2$$

My attempt : We can rewrite : $(x+10)^2-k^2 = 19x \Leftrightarrow (x+10-k)\cdot (x+10+k) = 19x$ But it doesn't seem to help...

Answer 1

$x^2+x +100 - k^2=0\implies \triangle = 1^2 - 4(1)(100-k^2) = 4k^2 - 399 = n^2 \implies (2k-n)(2k+n) = 399$. It seems that all the roads lead to ''$399$''.

Answer 2

How about $4x^2+4x+400=4k^2$?

$$(2k)^2-(2x+1)^2=399$$

$$(2k+2x+1)(2k-2x-1)=399$$

Answer 3

Notice that$$x^2 < x^2+x+100 <(x+10)^2.$$Thus $k=x+y$, where $1 \le y \le 9$. Then$$x^2+x+100=x^2+2xy+y^2 \Longrightarrow x=\frac{100-y^2}{2y-1}$$and$$2y-1\mid 100-y^2 \Longrightarrow 2y-1 \mid (2y-1)^2+4(100-y^2)=-4y+401.$$Finally $$2y-1 \mid 2(2y-1)-4y+401=399$$ and since $399=3\times7 \times 19$, hence $2y-1=1, 3, 7$.

Answer 4

$$4k^2=4x^2+4x+400$$ so $$(2k)^2-(2x+1)^2=399.$$ So $399=(2k+2x+1)(2k-2x-1)$ etc.

Answer 5

From DeepSea $\displaystyle (2k-n)(2k+n)=399=3.7.19\qquad x=\frac{-1\pm n}2$

Let's find all solutions in $\mathbb Z^2$.

$\begin{cases} 2k-n=a \\ 2k+n=b \end{cases}\iff \begin{cases} 4k=a+b \\ 2n=b-a \end{cases}$

Note that $\begin{array}{l} (a,b,k,n,x_1,x_2)\mapsto(-a,-b,-k,-n,x_2,x_1)\\ (a,b,k,n,x_1,x_2)\mapsto(b,a,k,-n,x_2,x_1)\\\end{array}$

So we need only to calculate for $0\le a\le b$ and take $\pm k$ as solution.

$\begin{array}{|cc|ccc|cc|} \hline a & b & a+b & b-a & n & k & x\\\hline 1 & 399 & 400 & 398 & 199 & \pm 100 & -100,99\\ 3 & 133 & 136 & 130 & 65 & \pm 34 & -33,32\\ 7 & 57 & 64 & 50 & 25 & \pm 16 & -13,12\\ 19 & 21 & 40 & 2 & 1 & \pm 10 & -1,0\\ \hline\end{array}$

The solutions in positive integers are $(0,10),(12,16),(32,34),(99,100)$