PHP hide divs on specific pages

Try this. If the GET parameter page isn't set or doesn't have a value, it will hide the div.

<?php
if(isset($_GET['page']) && $_GET['page'] != "") {
?>
<div class="icons_small"> 
<a href="#"><img src="images/layout/facebook_icon_s.png" width="75" height="67" /></a>
<a href="#"><img src="images/layout/twitter_icon_s.png" width="75" height="67" /></a>
<a href="#"><img src="images/layout/youtube_icon_s.png" width="75" height="67" /></a>
<a href="#"><img src="images/layout/location_icon_s.png" width="75" height="67" /></a>
</div>
<?php
}
?>

This would depend on how the pages are being included.

If we are using PHP's echo to print out the HTML, we could simply choose to not print that specific section out with echo when page is not set (that is, there is no page=whatever).

if(!isset($_GET['page'])
{
    echo 'The contents of the div that shouldn't be on the page';
}

If for whatever reason you cannot selectively print sections, a messy alternative would be to echo some JavaScript which would remove the div. However, this has the crucial flaw that the div would still be shown for users who have JavaScript disabled.

This is not secure at all. I understand it's all about 4 images but learning from a bad practice could be fatal for other projects.

What would happen if I write "index.php?page=I_WANT_TO_SEE_THAT_BOX" ?

What I would suggest is, check should you display it or not directly in your including function. I assume you have something like

function IncludeView() {
  $allowed_pages = array("view","friends","foo","bar");
  if(isset($_GET['page'])) {
    if(in_array($_GET['page'],$allowed_pages)) {
      include_once(PAGE_DIR."/".$_GET['page'].".php");  
      define("IS_PAGE_OKAY",true);
    }
  }
  else header(BASE_URL);
}
IncludeView();

Later in your template:

<?php if(defined("IS_PAGE_OKAY")) { ?>
   //your html code that is shown only on other pages than index.php
<?php } ?>