Does the squared error depends on the number of hidden layers?
Though this is not a programming question, I'll try my best to answer it here.
The squared error, i.e. the 'loss' of your neural network, depends on your neural network prediction and the ground truth. And it is convex from its definition.
The reasons that you're not getting low losses could be:

You're not normalizing your inputs. For example, if you got a series of house prices as input, which is around 500k to 1m, and you didn't normalize them, your prediction will be the linear combination of the prices, which is about the same order of magnitude, then pass through the activation function. This could result in large losses.
You're not initializing your weights and biases correctly. Similar to above, you could have large weights/biases which lead to large prediction values.
You didn't choose the proper activation function. When you're doing classification, your labels are generally one hot encoded, so your activation functions should limit the prediction to [0,1] or similar, so relu won't be a proper option. Also you don't want sigmoid as activation for regression problems.
Your labels are not predictable or have too much noise. Or maybe your network is not complex enough to capture important patterns, in that case you could try adding more layers and more nodes per layer.
Your learning rate is too small, this leads to slow convergence.

That's all I have in mind. You probably need more work to find out the reason to your problem.
I wanted to know if the squared error depends also on the number of hidden layers and the number of neurons on each hidden layer , because I've created a neuron network with one hidden layer but I can't reach a small squared error , so maybe the function is not convex ? Can I optimize weights by adding more hidden layers ?