Since it's a self-study problem, I'm just giving a few hints. Let the regression line of Y on X be $Y=a+bX$.

• That regression line goes through the point $(mean(X),mean(Y))$, therefore $mean(Y)=a+b·mean(X)$.
• Residuals $u_{X_i}$ $u_{Y_i}$ from the first part of your problem are just $X_i-mean(X)$ and $Y_i-mean(Y)$.
• By subtracting $Y=a+bX$ and $mean(Y)=a+b·mean(X)$ you can get $u_{Y_i}=a+b·u_{X_i}$ that is a regression line without intersect.
• If you need to show that this is the least squares regression of $u_{Y_i}$ on $u_{X_i}$, you can use that the residuals here will be the same than in $Y=a+bX$.

Since it's a self-study problem, I'm just giving a few hints. Let the regression line of Y on X be $Y=a+bX$.

• That regression line goes through the point $(mean(X),mean(Y))$, therefore $mean(Y)=a+b·mean(X)$.
• Residuals $u_{X_i}$ $u_{Y_i}$ from the first part of your problem are just $X_i-mean(X)$ and $Y_i-mean(Y)$.
• By subtracting $Y=a+bX$ and $mean(Y)=a+b·mean(X)$ you can get $u_{Y_i}=a+b·u_{X_i}$ that is a regression line without intersect.
• If you need to show that this is the least squares regression of $u_{Y_i}$ on $u_{X_i}$, you can use that the residuals here will be the same than in $Y=a+bX$.

Edit: "Regress $Y_i$ exclusively on the intercept", as I understand it, means fitting a model $Y=a$ (with $a$ constant). Adjusting this model using least squares yields just $Y=mean(Y)$.