We have 237=3+27=3+172=3+13+12=3+13+11+1237=3+27=3+172=3+13+12=3+13+11+1

So (a;b;c)=(3;3;1).

x3−4y=15x3−4y=15(1)

⇔xy−123y=15⇔xy−123y=15

⇔5xy−60=3y⇔5xy−60=3y

⇔5xy−3y=60⇔5xy−3y=60

⇔y⋅(5x−3)=60⇔y⋅(5x−3)=60

⇒y=605x−3⇒y=605x−3(2).

(1),(2) => x3−4605x−3=15⇔x3−20x−1260=15⇔x3−4⋅(5x−3)60=15⇔x3−5x−315=15⇔5x15−5x−315=15⇔315=15x3−4605x−3=15⇔x3−20x−1260=15⇔x3−4⋅(5x−3)60=15⇔x3−5x−315=15⇔5x15−5x−315=15⇔315=15 . 

From here we know that x has lots of value but just (x;y) = (1;30),(3;5) satisfy thread.

⇒5x=16+y3⇒5x=16+y3

⇒5x=16+2y6⇒5x=16+2y6

⇒5x=2y+16⇒5x=2y+16

⇒x(2y+1)=30⇒x(2y+1)=30

But 2y + 1 is a odd => x is a even number

=> x∈{−30;−10;−6;−2;2;6;10;30}x∈{−30;−10;−6;−2;2;6;10;30}

We have:

x -30 -10 -6 -2 2 6 10 30
2y+1 -1 -3 -5 -15 15 5 3 1
y -1 -2 -3 -8 7 2 1 0

So (x,y)=(−30;−1);(−10;−2);(−6;−3);(−2;−8);(2;7);(6;2);(10;1);(30;0)

⇒x6−130=2y⇒x6−130=2y

⇒5x30−130=2y⇒5x30−130=2y

⇒5x−130=2y⇒5x−130=2y

⇒y(5x−1)=60⇒y(5x−1)=60

But 5x−1≡4(mod5)5x−1≡4(mod5)

⇒5x−1∈{−6;−1;4}⇒5x−1∈{−6;−1;4}

If 5x - 1 = -6

=> 5x = -5 => x = -1

And y = 60 : -6 = -10

If 5x-1 = -1 

=> 5x = 0 => x = 0

And y = 60 : -1 = -60

If 5x-1 = 4 

=> 5x = 5 => x = 1

And y = 60 : 4 = 15

So (x,y)=(−1;−10);(0;−60);(1;15

C=10101⋅(5111111+5222222−43⋅7⋅11⋅13⋅17)C=10101⋅(5111111+5222222−43⋅7⋅11⋅13⋅17)

⇔C=511+522−148187⇔C=511+522−148187

⇒C=−41374⇒C=−41374