# Tag Info

## Hot answers tagged diophantine-equations

3

I could manage to find many almost solutions (with the aim to derive full solutions) by a systematic search: My first trials, code and ideas are posted here, where I received very helpful input that directed me to a promising approach, proposed by Peter: Search for 6-tuples $(s,t,u,t+u,t+u−s,t−s)$ consisting of perfect squares. I implemented a Python script ...

1

I am going from here: instead of dividing two, I am going to multiple two. $$4044=2((x-y)^2+(y-z)^2+(z-x)^2)=(2x-y-z)^2+3(y-z)^2$$ WLOG $x\ge y\ge z$ Therefore set $2x-y-z=a$ and $y-z=b$ we are finding the solution of $a^2+3b^2=4044$. After trying, we have three solutions: $(63,5),(39,29),(24,34)$. $(63,5)$ yields $x=34+z,y=5+z$, $(39,29)$ yields $x=34+z,y=... 1 No,there are no other positive integral solutions except the one you have already guessed. For this solution ,I would be replacing (n,m) with (x,y).(as I comfortable in solving with those variables)$\implies$$x=y^{x} \implies y=x^{1/x} By differentiating the expression,you will realise that the maximum value of the expression is at e which is e^{1/e}.... 1 For the case M=5, K=6, all solutions have x divisible by 6. Modulo 2, you'd get x^2\equiv 0\pmod{2}, so x is even. Modulo 3, you get x^2+y^2\equiv 0\pmod 3. A little work shows that this implies x must be divisible by 3. In fact, this shows that if x^2-20y^2 is divisible by 6 then x is divisible by 6. It really has nothing to do ... 1 x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0\iff(x^2+x)^2-4(x^2+x)=(y+1)^2+5 Putting X=x^2+x we get (X-2)^2=(y+1)^2+9 whose only solutions are clearly$$(X-2,y+1)=(\pm3,0),(\pm5,\pm4)$$in both cases, since X=x^2+x, we have the equations for the unknown x$$x^2+x-5=0\text{ and } x^2+x+1=0\\x^2+x-7=0\text{ and } x^2+x+3=0 these four equations have not integral ...

1

Since this is a Diophantine equation, we only seek for integer solutions. Notice that $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$ implies $(y-x^2-x+3)(y+x^2+x-1)=-9$, we only have $6$ cases: $((y-x^2-x+3),(y+x^2+x-1))$ must be one of $(1,-9),(9,-1),(3,-3),(-1,9),(-9,1),(-3,3)$. However, after checking all these cases there are no solutions. A simpler way to do this: ...

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