Maximum possible integer square root polynomial 1: easier (but still hard)

Let $y = \sqrt{25x^2-103x+2014}$ . Let $a = 10y, b= 50x-103$ . Then squaring , multiplying both sides by $100$ , and completing the square gives $(a+b)(a-b) = 190791$ .

Note that $a+b \equiv 7$ mod $10$ and $a-b \equiv 3$ mod $10$ . Note also that both factors should be positive because their sum is $20y$ , which is positive. There are six pairs of factors of $190791 = 3^2\cdot 17 \cdot 29 \cdot 43$ of this form: $63597 \cdot 3, 17 \cdot 11223, 4437 \cdot 43, 87 \cdot 2193, 1247 \cdot 153, 387 \cdot 493.$ This leads to the solutions $(x,y) =$ $(638,3180), (-110,562), (46,224), (-19,114), (13,70), (1,44).$ So the answer is $\fbox{3180}$ .

Let $k$ be an integer such that $25x^2-103x+2014=k^2$ $25x^2-103x+(2014-k^2)=0$ Since $x$ is an integer satifying the above equation, the discrminant of the above quadratic is a perfect square .

Let $p$ be that perfect square . $103^2-4(2014-k^2)(25)=p^2$ (expand it) $\Rightarrow 4(25)(2014)-103^2=(10k)^2-p^2$ $\Rightarrow (10k-p)(10k+p)=190791$

We know that $190791=3^2\times 17\times 29\times 43$

Let $\alpha_1$ and $\alpha_2$ be factors of $190791$ , such that

$(10k-p)=\alpha_1$ $(10k+p)=\alpha_2$ $\alpha_1 \alpha_2=190791$

Add the first 2 eqns

$20k=\alpha_1+\alpha_2$

So we need to find 2 factors of 190791 such that there sum is maximum, by AM-GM we know that the minimal value would occur at symmetry(both the factors being equal), so we can think that at the extreme points the sum would maximize. We start by taking one of the factors to be unity(one) and the other to be the 190791, but we fail to get an integer solution. Second we take one factor to be 3 and the other to be 63579, in this case we get an an integer solution i.e, $\boxed{k=3180}$

let $y=\sqrt{25x^2-103x+2014}$

Which represents the function $\sqrt{a^2x^2+bx+c}$

Thus, because the first term is obviously a square number for an integer X, $103x-2014$ must be a difference of squares.

We can represent $y^2=25(x+n)^2$ implying that $25(x+n)^2-25x^2=-103x+2014$

From here, a little manipulation yields $x=\dfrac{2014-25 n^2}{103+50 n}$

Now, I don't know of any guarnteed way of finding an exact n without guess and check. Luckily, however, the max value of x occurs when the denominator is as small as possible, or $103=-50n$

However, since n must be an integer for x to be an integer, we round to the nearest integer, thus n=-2. As our first guess, we can try n=-2, which, coincidentally, makes x an integer.

Thus plugging in n=-2, x=638. Plugging in x=638 we get $\boxed{y=3180}$

I will post a note for a much more generalized form of this.