Find integer $x\neq 0$ such that $120x+1,160x+1,240x+1$ are perfect square

Question

If $120x+1=a^2, 160x+1=b^2, 240x+1=c^2$ then we get $a^2+c^2=2b^2$ using $\pmod 2$ we get that $c+a=2u$, $c-a=2v$ for some $u,v$ so $a=u-v$ and $c=u+v$ then we get $(u-v)^2+(u+v)^2=2b^2$ or $u^2+v^2=b^2$ and from $a=u-v$, $c=u+v$ we could get that $30x=uv$ now since $160x+1=b^2$ is odd then between $u$ and $v$ one is odd and one is even and $uv$ must be the multiple of $30$. Now I dun know what I should do next. I try some Primitive Pythagorean triple but I dun seem to find any. Please help me and sorry for my bad latex arrangements

Edit: My approach above is wrong because of my mistake for thinking 120+240=2×160 so forgive me for this, I'm so sorry now I can't thinking of another approach... I hope someone could help me with this

Answer 1

Not an answer, but a thing that is too big for a comment.

I run some Mathematica code to look for solutions to your problem, using the following code:

In[1]:=ParallelTable[
 If[ArrayQ[{{Sqrt[120*x + 1]}, {Sqrt[160*x + 1]}, {Sqrt[
      240*x + 1]}}, _, IntegerQ], x, Nothing], {x, -10^6, 10^6}]

That code looks for answers on your question between the boundaries $-10^6\le x\le10^6$ and it gave me:

Out[1]={0}

So the only solution it found between the used boundaries is $x=0$.

Answer 2

Your setup says $$ 2a^2 - 3 b^2 + c^2 = 0 $$

All primitive solutions, $\gcd(a,b,c) = 1,$ come from $$ a = u^2 + 2uv - 2v^2 $$ $$ b = u^2 + 2 v^2 $$ $$ c = u^2 - 4uv - 2 v^2 $$ perhaps using the absolute values, with $u$ odd and $\gcd(u,v) = 1.$ Since your problem is primarily to do with $x$ it may be necessary to multiply, instead of such $a,b,c$ take $na,nb,nc$ for some positive $n$

You need