How can be done by the method of mathematical induction? We are given that $P(x+1)-P(x)=2x+1$
We also know that $P(0)=1$
We want to prove that $P(2004)=(2004)^2 +1$
Can someone explain how can be solved with mathematical induction?
Thank you in advance!

Observe that $P(y)=y^2+1$ satisfies the conditions [Proof below]
So, if $P(x)=x^2+1, P(x+1)=P(x)+2x+1=x^2+2x+1+1=(x+1)^2+1$

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Proof:
As $P(x+1)-P(x)=2x+1$
$P(x)$ can be at most Quadratic Polynomial
Let $P(x)=Ax^2+Bx+C$ as $P(0)=1, A\cdot0^2+B\cdot0+C+C\implies C=1$
So, $P(x)=Ax^2+Bx+1, P(x+1)=A(x+1)^2+B(x+1)+1$
So, $2x+1=P(x+1)-P(x)=A(2x+1)+B=2Ax+A+B$
Equating the coefficients of $x,2=2A\implies A=1$
Equating the constants $A+B=1\implies B=0$
]
