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I'm new to this site. Apparently It looks like that given $3$ positive integers $a,b,c$ such that $a>b>c$ it's always true that $$a^2 - b^2 > b^2-c^2$$

I'm having a hard time proving this. Thanks in advance.

Sahiba Arora
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ETHER
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4 Answers4

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That is pretty obvious. The left side is negative because b < a and the right side is positive because b > c.

Unless you didn't mean that.

EDIT to answer the edited question:

THAT is not true. Take 1, 3, 4. 16-9=7 while 9-1=8.

kkica
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  • Yeah :) ${}{}{}{}{}$ – Asinomás Jun 14 '17 at 16:17
  • mistake I have to edit the question – ETHER Jun 14 '17 at 16:18
  • I edited the answer – kkica Jun 14 '17 at 16:26
  • Thank you. my initial problem was to find a contradiction in equations series like p^2 + k = q^2, q^2 + k = t^2, t^2 + k = n^2 ....... where p,q,t, ... are positive integers and k is a positive integer constant to show that there can't be such k and I ended up asking above question. Can you give me any clue or should I ask a new question? – ETHER Jun 14 '17 at 16:37
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Under the new interpretation it is false.

Let $a=5,b=4,c=1$. Then $a^2-b^2=9$ and $b^2-c^2=15$.

Glorfindel
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Asinomás
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  • Got it, Thanks.. Yeah, Go it.. my initial problem was to find a contradiction in equations series like p^2 + k = q^2, q^2 + k = t^2, t^2 + k = n^2 ....... where p,q,t, ... are positive integers and k is a positive integer constant to show that there can't be such k and I ended up asking above question. Can you give me any clue or should I ask a new question? – ETHER Jun 14 '17 at 16:38
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Using $a>c$ we get $$a^2>ac>c^2$$ and $$-a^2<-c^2$$
Finally adding $b^2$.....

Or you could note that $$b^2-a^2<0<b^2-c^2$$

LM2357
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  • Thanks you for the answer.. I have typed the question wrong.. Just edited it – ETHER Jun 14 '17 at 16:20
  • But your new inequality is false..Try $a=10,b=9,c=1$ – LM2357 Jun 14 '17 at 16:24
  • Yeah, Go it.. my initial problem was to find a contradiction in equations series like p^2 + k = q^2, q^2 + k = t^2, t^2 + k = n^2 ....... where p,q,t, ... are positive integers and k is a positive integer constant to show that there can't be such k and I ended up asking above question. Can you give me any clue or should I ask a new question? – ETHER Jun 14 '17 at 16:35
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The thing you are asking about is false.

Take integers $r > s > 0$ with additional inequality $$ r > s \left( 1 + \sqrt 2 \right). $$ Then take $$ a = r^2 + 2rs - s^2, \; \; b = r^2 + s^2, \; \; c = r^2 - 2 r s - s^2. $$ The result is $$ a > b > c > 0, $$ $$ a^2 - b^2 = b^2 - c^2. $$ It is also true that $$ a-1 > b > c > 0, $$ $$ (a-1)^2 - b^2 < b^2 - c^2. $$

Will Jagy
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  • Yeah, Go it.. my initial problem was to find a contradiction in equations series like p^2 + k = q^2, q^2 + k = t^2, t^2 + k = n^2 ....... where p,q,t, ... are positive integers and k is a positive integer constant to show that there can't be such k and I ended up asking above question. Can you give me any clue or should I ask a new question? – ETHER Jun 14 '17 at 18:13
  • @MahelaGunawardana ask a new question – Will Jagy Jun 14 '17 at 18:32