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Question Number 205062 by BaliramKumar last updated on 07/Mar/24

	Answered by cortano12 last updated on 07/Mar/24

	$P_{\max} = 240$
	Commented by BaliramKumar last updated on 07/Mar/24

	$solution$
	Answered by mr W last updated on 07/Mar/24

	$y^{2} - x^{2} = 15^{2}$ $(y - x) (y + x) = 15^{2} = 1 \times 225$ ${(y - x)}_{m i n} = 1$ ${(y + x)}_{m a x} = 225$ $(x = 112, y = 113)$ $\Rightarrow P_{m a x} = {(y + x)}_{m a x} + 15 = 225 + 15 = 240$
	Commented by BaliramKumar last updated on 07/Mar/24

	$thanks$
	Answered by Rasheed.Sindhi last updated on 07/Mar/24
	$If a,b∈N with a>b then a^2 −b^2 , 2ab , a^2 +b^2 are sides of right triangle i-e pathagorean triple. 2ab is certainly an even number and a^2 +b^2 is hypotenuse. Hence a^2 −b^2 =15 (a−b)(a+b)=1×3×5 { ((a−b=1 ∧ a+b=15⇒(a,b)=(8,7))),((a−b=3 ∧ a+b=5⇒(a,b)=(4,1))) :} determinant ((a,b,(a^2 −b^2 ),(2ab),(a^2 +b^2 ),(perimeter)),(8,7,(15),(112),(113),(240)),(4,1,(15),8,(17),(40))) The above formula works only for primitive triples.To find out non- primitive triples we can use this same formula to find out primitive triples, first, with one side 3 or 5: a^2 −b^2 =3⇒(a−b)(a+b)=1×3 a−b=1∧ a+b=3⇒(a,b)=(2,1) a^2 −b^2 =5⇒(a−b)(a+b)=1×5 a−b=1∧ a+b=5⇒(a,b)=(3,2) determinant ((a,b,(a^2 −b^2 ),(2ab),(a^2 +b^2 )),(2,1,3,4,5),(3,2,5,(12),(13))) (3,4,5)×5=(15,20,25)⇒p=60 (5,12,13)×3=(15,36,39)⇒p=80 Yet the max p=240$
	$I f a, b \in N w i t h a > b t h e n$ $a^{2} - b^{2}, 2 a b, a^{2} + b^{2} a r e s i d e s o f$ $r i g h t t r i a n g l e i - e p a t h a g o r e a n$ $t r i p l e .$ $2 a b i s c e r t a i n l y a n e v e n n u m b e r$ $a n d a^{2} + b^{2} i s h y p o t e n u s e . H e n c e$ $a^{2} - b^{2} = 15$ $(a - b) (a + b) = 1 \times 3 \times 5$ ${\begin{cases} a - b = 1 \land a + b = 15 \Rightarrow (a, b) = (8, 7) \\ a - b = 3 \land a + b = 5 \Rightarrow (a, b) = (4, 1) \end{cases}$ $\begin{array}{ccc} a & b & a^{2} - b^{2} & 2 a b & a^{2} + b^{2} & p e r i m e t e r \\ 8 & 7 & 15 & 112 & 113 & 240 \\ 4 & 1 & 15 & 8 & 17 & 40 \end{array}$ $The above formula works only for$ $primitive triples . To find out non -$ $primitive triples we can use this$ $same formula to find out primitive$ $triples, first, with one side 3 or 5 :$ $a^{2} - b^{2} = 3 \Rightarrow (a - b) (a + b) = 1 \times 3$ $a - b = 1 \land a + b = 3 \Rightarrow (a, b) = (2, 1)$ $a^{2} - b^{2} = 5 \Rightarrow (a - b) (a + b) = 1 \times 5$ $a - b = 1 \land a + b = 5 \Rightarrow (a, b) = (3, 2)$ $\begin{array}{ccc} a & b & a^{2} - b^{2} & 2 a b & a^{2} + b^{2} \\ 2 & 1 & 3 & 4 & 5 \\ 3 & 2 & 5 & 12 & 13 \end{array}$ $(3, 4, 5) \times 5 = (15, 20, 25) \Rightarrow p = 60$ $(5, 12, 13) \times 3 = (15, 36, 39) \Rightarrow p = 80$ $Y e t t h e m a x p = 240$
	Commented by BaliramKumar last updated on 07/Mar/24

	$thanks$ $i think 4 integral side triangles possible$
	Commented by Rasheed.Sindhi last updated on 07/Mar/24

	$Y e s s i r, I^{'} v e e x t e n d e d m y a n s w e r t o$ $c o v e r a l l p o s s i b l e t r i p l e s n o w .$
	Commented by BaliramKumar last updated on 07/Mar/24

	$Yes sir$ $also y^{2} - x^{2} = 15^{2} = 225$ $(y - x) (y + x) = 1 \times 225$ $(y - x) (y + x) = 3 \times 75$ $(y - x) (y + x) = 5 \times 45$ $(y - x) (y + x) = 9 \times 25$ $(y - x) (y + x) = 15 \times 15$
	Commented by Rasheed.Sindhi last updated on 08/Mar/24

	$A n e a s y w a y!$
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