Is there an integer solution to $x^2+1978=y^2$

Question

Is there an integer solution to $x^2+1978=y^2$?

Don't know really how to approach this. Thanks

Answer 1

HINT:

$$(y+x)(y-x)=1978\equiv2\pmod4$$

But $y+x+(y-x)=2y$ or $y+x-(y-x)=2x$ is even $\implies y\pm x$ have the same parity

Answer 2

No. Squares are congruent to $0$ or $1$ modulo $4$, but $1978 \equiv 2 \mod 4$.

Answer 3

$(y+x)(y−x)=1978$

$1978=2\times23\times43$

is (i) $46\times43=1978$ or;
(ii)$23\times86=1978$

Case(i) $$x+y=46$$ $$y-x=43$$ $$\Longrightarrow x=1.5 \wedge y=44.5$$ Case (ii) $$x+y=86$$ $$y-x=23$$ $$\Longrightarrow x=31.5 \wedge y=54.5$$

Which means the equation have no integer solution

Answer 4

An integer number $n$ is represented by the quadratic form $x^2-y^2$ iff $n\not\equiv 2\pmod{4}$, since $$ x^2-y^2 = n $$ is equivalent to: $$ (x-y)(x+y)=n, $$ that can be read as "$n$ has two complementary divisors $d,\frac{n}{d}$ with the same parity".