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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

user635988
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    Assumedly you want $x \neq 0$ – Good Boy Jun 03 '20 at 15:37
  • @GoodBoy I forget to put that, my bad – user635988 Jun 03 '20 at 15:44
  • No worries: also, do you not get $$2 a^2=c^2+1,$$ by subtracting the third equation from two times the first? (and two other equations) – Good Boy Jun 03 '20 at 15:45
  • @GoodBoy ohhh I totally didn't notice that. I only focus on the arithmetic progress – user635988 Jun 03 '20 at 15:48
  • $120+240$ is not the same as $2\cdot 160$. – cr001 Jun 03 '20 at 15:57
  • @cr001 ohh what happened to me thinking it is right...sorry sir... – user635988 Jun 03 '20 at 16:01
  • Now everything is a mess should I edit it? – user635988 Jun 03 '20 at 16:02
  • $$ 2 a^2 - 3 b^2 + c^2 = 0 $$ – Will Jagy Jun 03 '20 at 17:59
  • The $x$ must be a multiple of $7$. – Vepir Jun 03 '20 at 18:15
  • $c^2-2a^2=-1$ so by the theory of Pell equations $c+a\sqrt2=(1+\sqrt2)^t$ with $t$ odd. Getting $b$ involved by another Pell equation involving $2a$ and $b$. The solution sets of those two Pell equations are so sparse that a collision feels unlikely. – Jyrki Lahtonen Jun 03 '20 at 18:38
  • Unless I made a mistake we need $4a$ to be equal to $\sqrt2[(1+\sqrt2)^{2t+1}-(1-\sqrt2)^{2t+1}]$ for some natural number $t$ and also equal to $(2+\sqrt3)^\ell-(2-\sqrt3)^\ell$ and we need $a$ to $\equiv\pm1$ or $\pm11\pmod{30}$. The last is not serious for it is satisfied by two thirds of the choices of $t$. Matching $t$ and $\ell$ on the other hand... May be there is an argument involving units of $\Bbb{Q}(\sqrt2,\sqrt3)$? – Jyrki Lahtonen Jun 03 '20 at 19:05

2 Answers2

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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 $x=0$.

Jan Eerland
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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

Will Jagy
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