How many integer sided triangles (i.e. all the three sides are positive integers) are there such that they have the same area and the perimeter (i.e. they are numerically same notdimensionally)?

I found that the answer is that THERE IS NO SUCH TRIANGLE and I had proved it , but I doubt mysolution is correct.Please help me.

Dear Adhiraj,
I would like to see your proof.
PLease Post it.
Regards.

please post your proof.

Let ABC be one such triangle with side lengths a,b,and c

Then a+b+c=sq.rt.[s(s-a)(s-b)(s-c)][By HERON''S FORMULA]

so,(a+b+c)(a+b+c)={(a+b+c)/2}{(b+c-a)/2}{(a+c-b)/2}{(a+b-c)/2}[putting the value of s and calculating]

(a+b+c)^2={(a+b+c)/2}{(b+c-a)/2}{(a+c-b)/2}{(a+b-c)/2}

(a+b+c)^2=1/4[(b+c)^2-(a^2)]*1/4[(a^2)-(b-c)^2]

(a+b+c)^2 =1/16[b^2+c^2+2bc-a^2][a^2-b^2-c^2+2bc]

(a+b+c)^2 =1/16[{(2bc)^2}-{(b^2+c^2-a^2)^2}]

(a+b+c)^2 =1/16[4b^2c^2-(b^4+c^4+a^4+2(b^2c^2)-2(c^2a^2)-2(b^2a^2)]

(a+b+c)^2 =1/16[4(b^2c^2)-b^4-c^4-a^4-2(b^2c^2)+2(c^2a^2)+2(b^2a^2)]

(a+b+c)^2 =1/16[2(b^2c^2)-b^4-c^4-a^4+2(c^2a^2)-2(b^2a^2)]

(a+b+c)^2 =-1/16[a^4+b^4+c^4-2(b^2c^2)-2(c^2a^2)+2(b^2a^2]

(a+b+c)^2 =-1/16[(a^2+b^2-c^2)^2]

Therefore,POSITIVE=NEGATIVE......................HENCE CONTRADICTION

However,is ther any flaw?I do not know .There may be,but I can not see one.