How many Integer values of $n$ are possible for $n^2+25n+19$ to be a perfect square.

Question

[1] How many Integer values of $n$ are possible for $n^2+25n+19$ to be a perfect square.

[2] How many Integer values of $n$ are possible for $n^2-19n+99$ to be a perfect square.

$\underline{\bf{My\;Try}}::$ for first one , Let $k^2 = n^2+25n+99$, where $k,n\in \mathbb{Z}$

So $4k^2 = 4n^2+100n+76\Rightarrow (2k)^2 = (2n)^2+2\cdot (2n)\cdot 25+625+(76-625)$

$(2k)^2 = (2n+25)^2-549\Rightarrow (2n+25)^2-(2k)^2 = 549 = 3^3\cdot 61$

Now Let $x= 2k$ and $y = (2n+25)$,

we get $(x^2-y^2)=(x+y)\cdot(x-y) = 3^2 \cdot 61$

Is it Right or not ,and is there is any other method to solve these type of questions,

If Yes the please explain here

Thanks

You could also try to solve the quadratic equation n^2 + 25n + 19 - k^2 = 0, which will get you again around this magic 549 number ;) — Cristi, Oct 13 '13 at 10:20

lab bhattacharjee · Answer 1 · 2013-10-13T10:14:03.193

6

Let $$n^2+25n+19=(n+a)^2\text{ where }a\text{ is some integer }$$

$$\iff n=\frac{a^2-19}{25-2a}$$

Let integer $d$ divides both $a^2-19,25-2a$

$\implies d$ divides $\{2(a^2-19)+a(25-2a)\}=25a-38$

$\implies d$ divides $\{2(25a-38)+25(25-2a)\}=549=9\cdot61$

So, the divisors$(d)$ of $549$ are $\pm1,\pm3,\pm9\pm61,\pm183,\pm549$

As $25-2a$ must divide $a^2-19,$ check which values of $d(=25-2a)$ make $n$ integer

edited Oct 13 '13 at 10:14

answered Oct 13 '13 at 09:54

lab bhattacharjee

you forgot divisor 183 ;) – Cristi Oct 13 '13 at 10:13
@Cristi, I generally write in pair like $\pm1,\pm649;\pm3,\pm183;\pm9,\pm61$ to avoid omission.Thanks anyway – lab bhattacharjee Oct 13 '13 at 10:15
Thanks lab bhattacharjee But I did not understand Let integer $d$ divides both $a^2-19,25-2a$
$\implies d$ divides ${2(a^2-19)+a(25-2a)}=25a-38$
– juantheron Oct 13 '13 at 10:57
@juantheron, if $d$ divides $a,b;d$ will divide $xa+yb$ where $x,y$ are integers, right? – lab bhattacharjee Oct 13 '13 at 10:59

1 Answers1