$P(-1)=1-a+b, P(0)=b , P(1)=1+a+b$ Given $(1-a+b)^2$,$b^2$ ,$(1+a+b)^2$ are A.P. of positive integers thus required to show $ (1-a+b)^2+(1+a+b)^2= 2b^2$ this equation reduces to $1+a^2+2b =0$,.................(1) thus we can say from here that b is negative and we know that $(1-a+b)^2$ is an integer from this we can say $1+a^2+b^2+2b-2a-2ab=b^2-2a-2ab$ is an integer Since $b^2$ is an integer given thus $2a+2ab$ is also an integer we also know that $(1+a+b)^2$ is an integer. Using 1 again, we can say $1+a^2+b^2+2b+2a+2ab=-2b+b^2+2a+2b+2ab$ is an integer Again using 1 we can say $b^2+2a-a(1+a^2)$ is an integer This implies $(a-a^3)=t $(assume) is an integer if b^2=c, then $ b= - \sqrt c$ $1+a^2= 2 \sqrt c$ then $ 2a(1- \sqrt c)=t$ Sqaring both sides we get $ \sqrt c=(t^2+20 c+4)/4(4+2c)$ Since RHS is rational, LHS is also rational, hence $\sqrt c$ is rational , since c is an integer , $\sqrt c= b$ is also an integer. now $a^2=-2b-1$ is also an integer since $a(1-a^2)$ is an integer, $a$ is also an integer Thus PROVED

How many marks will one get for writting till the part when $2a(1 - \sqrt{c})$ is an integer

SAUDITYA wrote: How many marks will one get for writting till the part when $2a(1 - \sqrt{c})$ is an integer I did the next part just to prove that b is an integer and only after getting b as an integer I proved a to be an integer as well. So in my proof the later part was a must, but I hope partial markings <= 5 ( I feel) can be given

I have got a simpler solution to this one.. The first part is somewhat similar to @i_atmmaths but the end is different.. here it is - Clearly, $a+ab\in\Bbb{Q}$ ........$(\alpha)$ And, $a^2=-2b-1$ ........$(\beta)$ We know, $b^2\in\Bbb{Z}$ Let, $b^2=k ,$ where $k\in\Bbb{Z}$ $\therefore$ $b$ must be $-\sqrt{k}$ (see $\beta$) $\alpha^2$ : $a^2+a^2b^2+2a^2b\in\Bbb{Q}$ Putting $b=-\sqrt{k}$ and $a^2=2\sqrt{k}-1$ we get that - $2\sqrt{k}-1+(2\sqrt{k}-1)(-\sqrt{k})+2(2\sqrt{k}-1)(-\sqrt{k})\in\Bbb{Q}$ After simple calculation, we find $5\sqrt{k}\in\Bbb{Q}$ $\implies \sqrt{k}\in\Bbb{Q}$ But, $k\in\Bbb{Z}$ Which clearly implies that $\sqrt{k}\in\Bbb{Z}$ $\therefore$ $b\in\Bbb{Z}$ Now, $\because a+ab\in\Bbb{Q}$ $\therefore a\in\Bbb{Q}$ But, $a^2\in\Bbb{Z}$ $\implies a\in\Bbb{Z}$ Thus we can conclude that both $a$ and $b$ must belong to $\Bbb{Z}$ This completes the proof.. Q.E.D. $\square$

div5252 wrote: 2.Let $P(x) = x^2 + ax + b$ be a quadratic polynomial where a, b are real numbers. Suppose $P(-1)^2$ , $P(0)^2$, $P(1)^2$ is an Arithmetic progression of positive integers. Prove that a, b are integers. As $p(0)^2=b^2=Z\in N$ so $b=\pm\sqrt{z}$ Now as $p(-1)^2, p(0)^2, p(1)^2$ are in AP so we have $a^2+1+2b=0 \implies a^2=-2b-1$ as $a\in R$ so $b=-\sqrt{z}$ and $a=\sqrt{2\sqrt{z}-1}$ also as $(a+b+1)^2=z+2(2\sqrt{z}+\sqrt{z(2\sqrt{z}-1)}+\sqrt{2\sqrt{z} -1})=x \in Z$ and we also have $(1+b-a)^2=z+2(2\sqrt{z}-\sqrt{z(2\sqrt{z}-1)}-\sqrt{2\sqrt{z} -1})= y\in Z$ Now by adding above $2$ equation we get $8\sqrt{z}\in Z \implies z=k^2, b=-k$ Now by subtracting those $2$ equations we get $\sqrt{2k-1}(k+1)\in Z \implies \sqrt{2k-1}=a\in Z$ hence $a, b$ are integers $\blacksquare$

@i_atmmaths how u knew that b² is the middle term

@above By putting in 0 = x and squaring