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Question:

Are there $a,b,c,d \in \mathbb N$ such that $$a^2 + b^2 = c^2 \ \ \text{and} \ \ b^2 + c^2 = d^2$$

I'm a bit lost here.

Martin Sleziak
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athos
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  • How about $a=b=c=d=0$? – Regret Feb 13 '15 at 11:39
  • @Regret let's say by $\mathbb N$ here means $a,b,c,d > 0$. – athos Feb 13 '15 at 11:43
  • your condition gives $a^2+2b^2 = d^2$ and I think someone asked this question on MSE for not so long. – Surb Feb 13 '15 at 12:00
  • The standard formulas for Pythagoras triples gives, $a=n^2-m^2,b=2mn=2pq,c=n^2+m^2=p^2-q^2,d=p^2+q^2$, so $(p\pm q)^2-2q^2$ must both be square numbers, equal to $(m\pm n)^2$. – Empy2 Feb 13 '15 at 12:28
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    The votes to close say, "This question is not about math..." Ah, I guess this question is about the letters of the alphabet. – Tito Piezas III Feb 13 '15 at 12:46
  • @Michael I agree with you till $(p+q)^2-2q^2=(m+n)^2$ and $(p-q)^2-2q^2=(m-n)^2$... but how could I proceed from here? – athos Feb 15 '15 at 15:15
  • See also: http://math.stackexchange.com/questions/960816/are-there-any-positive-integers-a-b-c-d-such-that-both-a-b-c-and-b – Martin Sleziak Feb 15 '15 at 17:07
  • @TitoPiezasIII If you have a closer look, the people who cast the first two close votes chose the following close reason: "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin Sleziak Feb 15 '15 at 17:09
  • @MartinSleziak: When it still had three close votes, I looked at and the tick mark was only on "Off-topic because... This question does not appear to be about math..." – Tito Piezas III Feb 16 '15 at 01:28
  • @TitoPiezasIII That seems strange. I reported what I saw after the first two closed votes. Then I case a close vote to close as a duplicate. Anyway, both revision history and close review only show who voted and what was the reason chosen by most of the users. I don't see any way to find out now, after the question was already closed, what the particular users voter for. – Martin Sleziak Feb 16 '15 at 06:44
  • @TitoPiezasIII But I think that the reason for what you see is that various close reasons are lumped together under off-topic. The description of off-topic is: This question does not appear to be about math within the scope defined in the help center. But when you choose to close off-topic, then you select one of the more specific reasons, one of them is the one I mentioned above. (You can try this on any question which is not closed. Simply click on close, then click on off-topic to see the choices offered to you. Of course, you should not proceed further to vote, if it is just a test.) – Martin Sleziak Feb 16 '15 at 06:47

2 Answers2

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Note that $c^2-b^2=a^2$, $c^2+b^2=d^2$ $\Rightarrow$ $c^4-b^4=(ad)^2$. But here (Wiki) and here (MSE) one can read that equation $$x^4-y^4=z^2$$ has no (pairwise coprime) solutions $x,y,z\in \mathbb{N}$.

Community
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Oleg567
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Suppose there exist such $a,b,c,d$ with $a\le b\le c\le d,$ then $$c^2-a^2=d^2-c^2=b^2$$ $$2c^2=a^2+d^2$$ Suppose $$a=x-y,\,\,\,\,d=x+y.$$ Then $$c^2=x^2+y^2.$$ Therefore there exist $A,B$ such that $$c=A^2+B^2,\,\,\,x=A^2-B^2,\,\,\,y=2AB.$$ Now we have $$a=A^2-2AB-B^2\le(A-B)^2$$ $$d=A^2+2AB-B^2\le(A+B)^2$$ Also $$c^2-a^2=(A^2+B^2)^2-(A^2-2AB-B^2)^2=4AB(A-B)(A+B)$$ should be a PERFECT SQUARE.
That is all $A,B,A+B,A-B$ are should be perfect squares. Therefore $$A=a_1^2,\,\,\,B=b_1^2$$

This happens only if $B$ is a perfect square and $A=0.$ Hence the only solution is $$a=c=d,\,\,\,\,\,b=0.$$

Bumblebee
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  • nice answer (+1) – Shobhit Feb 13 '15 at 12:38
  • This last looks like a restatement of the original question, with $A,B$ replacing $b^2,c^2$. Does this show that any solution with $a,b,c,d$ gives a smaller solution with $A,B,...$? – Empy2 Feb 13 '15 at 12:43
  • I also notice that.I'm try to fixed it. Do you have any idea? – Bumblebee Feb 13 '15 at 12:45
  • A common method in this area of maths is to show there is always a smaller solution. Since it can't go on forever, that means it can't start - so there is no solution in the first place. – Empy2 Feb 13 '15 at 12:54