No Integral Solution for x & y when c Not Divisible by d

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Homework Statement


For, a, b, c\inintegers and D\inintegers -{0}, if a and b are divisible by d, and c is not divisible by d then the equation ax+by=c has no integral solution for x and y.


Homework Equations





The Attempt at a Solution


ax+by=c
a/dx+b/dx=c
1/d[ax+by]=c
ax+by=cd
 
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Where did the d in a/d come from? You can't just "stick" it into part of the equation.

Saying "d divides a" means a= dn for some integer n. Saying "d divides b" means b= dm for some integer m. Replace a and b in your equation by that and see what happens.
 
HallsofIvy said:
Where did the d in a/d come from? You can't just "stick" it into part of the equation.

Saying "d divides a" means a= dn for some integer n. Saying "d divides b" means b= dm for some integer m. Replace a and b in your equation by that and see what happens.

dnx+dmx=c
d(nx+mx)=c
 
I don't know if that was the correct way to do it and then maybe say since we don't have a dpc then there is no integral solution?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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