ch2.0.14-questions-squares

Chapter XIV — Solution of some Questions that belong to this part of Algebra

Summary: Seventeen worked Diophantine showcase problems that tie together every artifice introduced in Part II so far: rationalizing surds, bootstrap from a seed, residue-class impossibility, Pell composition, and infinite descent. The chapter is essentially a problem set rather than new theory, but two general theorems emerge — that $a+x$ and $a-x$ can be both squares only when $a$ is a sum of two squares, and that $p^2+q^2$ and $p^2+3q^2$ cannot both be squares.

Sources: chapter-2.0.14

Last updated: 2026-05-09

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Orientation (§212)

Euler announces that the chapter contains no new artifices but illustrates those already explained “by adding here some of those questions with their solutions.” The seventeen questions span all of Part II’s techniques applied to formulas that must become squares.

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Q1 — $x\pm 1$ both squares (§213)

Set $x+1=p^2$, so $x-1=p^2-2$ must also be a square; let it be $(p-q{)}^2$. Then $p^2-2=p^2-2pq+q^2$, giving $p=(q^2+2)/(2q)$, so

$x=\frac{q^4+4}{4q^2}.$

Substituting fractional $q=r/s$ yields the family $x=(r^4+4s^4)/(4r^2{s}^2)$. Sample table:

$r,s$	$1,1$	$2,1$	$1,2$	$3,1$	$4,1$
$x$	$5/4$	$5/4$	$65/16$	$85/36$	$65/16$

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Q2 — $x+4,x+7$ both squares (§214)

Same method: $x+4=p^2$ gives $x=p^2-4$; then $x+7=p^2+3$ must be a square $(p+q{)}^2$, yielding

$p=\frac{3-q^2}{2q},x=\frac{9-22q^2+q^4}{4q^2}.$

With $q=r/s$: $r=s=1$ gives $x=-3$; $s=2,r=1$ gives $x=57/16$ (so $x+4=121/16$, $x+7=169/16$); $s=3,r=1$ gives $x=133/9$.

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Q3 — $1\pm x$ both squares (§215)

Setting $1+x=p^2$, $x=p^2-1$, the second formula $1-x=2-p^2$ must be a square — but no obvious seed except $p=1$ (the trivial $x=0$). Bootstrap with $p=1-q$: gives

$q=\frac{2r+2}{r^2+1},x=\frac{4r-4r^3}{(r^2+1{)}^2}.$

With $r=t/u$: $x=4tu(u^2-t^2)/(t^2+u^2{)}^2$. Need $u>t$. Examples: $u=2,t=1$ gives $x=24/25$; $u=3,t=2$ gives $x=120/169$ (so $1+x=289/169$, $1-x=49/169$).

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Q4 — $10\pm x$ both squares (§216), and the §217 Remark

Different mode: the product $100-x^2$ must be a square; set $100-x^2=(10-px{)}^2$ to get $x=20p/(p^2+1)$. Then $10+x=10(p+1{)}^2/(p^2+1)$, and reducing the leftover $10/(p^2+1)$ to a square requires $10p^2+10$ to be a square, with seed $p=3$. Bootstrap with $p=3+q$ produces

$x=\frac{20p}{p^2+1}\text{with}p=3+q,q=(60-20t)/(t^2-10).$

Selected values: $t=3$ gives $x=6$ ($10+x=16$, $10-x=4$); $t=1$ gives $x=-234/25$ (so $|x|=234/25$, and $10\pm x=484/25,16/25$).

Remark §217-218: when is the question solvable for general $a$?

Replace $10$ by an arbitrary $a$. The question is impossible whenever $a$ is not a sum of two squares. Reason: if $a+x=p^2$ and $a-x=q^2$, addition gives $2a=p^2+q^2$, so $2a$ is a sum of two squares — and (Article 170) $a$ must be too. The list of $a\le 50$ that fail (cannot be expressed as $x^2+y^2$):

$3,6,7,11,12,14,15,19,21,22,23,24,27,28,30,31,33,35,38,39,42,43,44,46,47,48.$

For these $a$, no $x$ — fractional or integer — can make both $a\pm x$ squares.

For $a=3$ specifically: would need $p^2/q^2+r^2/s^2=3$, i.e., $3q^2{s}^2=p^2{s}^2+q^2{r}^2$, the right side a sum of two squares divisible by 3. By Article 170 a sum of coprime squares can have no divisor that is not itself a sum of two squares — and 3 is not. Hence impossible.

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Q5 — Resolving a sum of two squares another way (§219)

Given $f^2+g^2$, find $x^2+y^2=f^2+g^2$ with new $x,y$. Set $x=f+pz$, $y=g-qz$; the cancellation yields

$z=\frac{2gq-2fp}{p^2+q^2},x=\frac{2gpq+f(q^2-p^2)}{p^2+q^2},y=\frac{2fpq+g(p^2-q^2)}{p^2+q^2}.$

Free parameters $p,q$. Example with $f=g=1$ (so the target is 2) and $p=2,q=1$: $x=1/5,y=7/5$, giving $x^2+y^2=2$.

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Q6 — $a\pm x$ both squares when $a=c^2+d^2$ (§220-221)

Concrete case $a=13=9+4$: set $13+x=p^2$, $13-x=q^2$, so $p^2+q^2=26$, also a sum of two squares ($25+1$). Quick solution $p=5,q=1$ gives $x=12$. For an infinite family, parametrize via Q5 with $f^2+g^2=26$: writing $t,u$ for the new parameters, $p=(2tu+5(u^2-t^2))/(t^2+u^2)$ etc., yielding $x=(p^2-q^2)/2$. Example $t=2,u=1$: $x=204/25$.

General case $a=c^2+d^2$, write $z$ for the unknown: parameters $p=1,q=1$ recover the trivial $x=2cd$, $y=c+d$, i.e., $a+z=(c+d{)}^2$ and $a-z=(c-d{)}^2$. With $p=2,q=1$ a second formula

$x=\frac{c+7d}{5},y=\frac{7c-d}{5},z=\frac{24d^2+14cd-24c^2}{25}$

generates further pairs.

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Q7 — $x+1$ and $x^2+1$ both squares (§222)

Set $x+1=p^2$, so $x=p^2-1$ and $x^2+1=p^4-2p^2+2$ must be a square. The seed $p=1$ (trivial $x=0$) yields nothing; bootstrap $p=1+q$ produces $x^2+1=1+4q^2+4q^3+q^4$. Five different root-ansätze on this quartic each peel off a subset of terms:

Root ansatz	$q$	$p$	$x$
$1+q^2$	$-1/2$	$1/2$	$-3/4$
$1-q^2$	$-3/2$	$-1/2$	$-3/4$
$1+2q+q^2$	$-2$	$-1$	$0$
$1-2q-q^2$	$-2$	(same)	(same)
$1+2q^2$	$4/3$	$7/3$	$40/9$

The $p=7/3,x=40/9$ case gives $x+1=49/9=(7/3{)}^2$ and $x^2+1=1681/81=(41/9{)}^2$. Iterating from $q=-1/2+r$ gives a further $r=21/20$, $x=561/400$.

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Q8 — $x+a$, $x+b$, $x+c$ all squares (§223-224)

Setting $x+a=z^2$, the three formulas reduce to making $z^2$, $z^2+(b-a)$, $z^2+(c-a)$ all squares. For most $b-a,c-a$ this is impossible, “and its possibility entirely depends on the nature of the numbers $b-a$ and $c-a$.”

Five impossibility cases shown by reduction to known no-square theorems:

• $b-a=1$, $c-a=-1$: would need $p^4-q^4$ a square — impossible by ch2.0.13-impossibility-biquadrate-sums.
• $b-a=2$, $c-a=-2$: $p^4-4q^4$ square — impossible (Article 209).
• $m=f^2$, $n=-f^2$: difference of two biquadrates — impossible.
• $m=2f^2$, $n=-2f^2$: $p^4-4r^4$ — impossible.
• $m=1$, $n=2$: forces $r^4-q^4$ to be a square.

Method to find solvable $(m,n)$ directly: start with arbitrary $f,g,h,k$, set $m=(h^2-f^2)/g^2$, $n=(k^2-f^2)/g^2$, so a seed $p=f,q=g$ makes both formulas squares automatically. Example $h=3,k=5,f=1,g=2$ yields $m=2,n=6$: $p^2+2q^2$ and $p^2+6q^2$ to be squares, with seed $p=1,q=2$. Bootstrap then produces $p^2+2q^2=191^2\cdot 2=43681=209^2$ and $p^2+6q^2=58081=241^2$ for $p=191,q=60$ (after some manipulation).

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Q9 — General $p^2+az{q}^2$ and $p^2+bz{q}^2$ both squares (§225-228)

Setting $p^2+az{q}^2=r^2$, $p^2+bz{q}^2=s^2$: subtract scaled versions, get $p^2=(b{r}^2-a{s}^2)/(b-a)$, which must be a square. Setting $r=s+(b-a)t$ and $p=s+xt/y$ leads (after lengthy calculation) to

$z=\frac{4xy((b-a)y-x)(by-x)}{q^2},p=(x-by{)}^2-ab{y}^2.$

After substituting $x=v+by$ for simplification: $p=v^2-ab{y}^2$ and $z=4vy(v+ay)(v+by)/q^2$, with $r=-v^2-2avy-ab{y}^2$, $s=-v^2-2bvy-ab{y}^2$.

Example 1 (§226): $a=-1,b=1$ — i.e., $p^2-z{q}^2$ and $p^2+z{q}^2$ both squares

$p=v^2+y^2$ and $z=4vy(v-y)(v+y)/q^2$. Tabulation:

$v,y$	$z$	$p$
$2,1$	$6$	$5$
$3,2$	$30$	$13$
$4,1$	$15$	$17$
$5,4$	$5$	$41$
$16,9$	$7$	$337$
$8,1$	$14$	$65$

Hence $p^2\pm 6q^2$ both squares at $p=5,q=2$: $25\pm 24=\{1,49\}$. Likewise $\pm 30q^2$ at $p=13,q=2$: $169\pm 120=\{49,289\}$.

Example 3 (§228) — Theorem §229-230: $p^2+q^2$ AND $p^2+3q^2$ never both squares

A negative result: Euler proves rigorously that the system has no solution. Sketch:

1. $\gcd (p,q)=1$ WLOG.
2. $p$ must be odd (else $p^2+3q^2\equiv 4n+3$, not a square).
3. Then $q$ divisible by 4 (so $q^2\equiv 0\mathrm{mod}16$, giving $p^2+q^2\equiv 8n+1$).
4. $p$ not divisible by 3, so $p^2\equiv 1\mathrm{mod}3$.
5. $q$ divisible by 3.
6. $p$ not divisible by 5; $p^2\equiv 1$ or $4\mathrm{mod}5$, leading to $q$ divisible by 5.
7. Hence $q=60m$, so $7200m^2=s^2-r^2=(s+r)(s-r)$ with $s,r$ odd coprime.
8. $r=f-g$ with $f,g$ coprime, opposite parity, $fg=1800m^2$.
9. With $m=1$: $fg=1800$, four candidate $(f,g)$ pairs all yield $r\equiv 4n+3$ — impossible since $r$ must equal $f-g=$ sum of two squares ($\equiv 4n+1$).
10. Apply infinite descent: any small solution would be smaller, contradiction.

The proof’s structure: residue classes prune $(p,q)$ to $q=60m$, then a divisor/parity analysis rules out every residue class for $r$.

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Q10 — $xy+1$, $xz+1$, $yz+1$ all squares (§231-232)

“Very elegant solution”: set $xz+1=p^2$, $yz+1=q^2$. Then $x=(p^2-1)/z$, $y=(q^2-1)/z$, and the third formula

$xy+1=\frac{(p^2-1)(q^2-1)}{z^2}+1$

must be a square. Multiplying by $z^2$: $(p^2-1)(q^2-1)+z^2=$ square; with root $z+r$, get

$z=\frac{(p^2-1)(q^2-1)-r^2}{2r}.$

Free $p,q,r$. With $r=pq+1$: $z=(p^2+2pq+q^2)/(2pq+2)=(p+q{)}^2/(2(pq+1))$, giving

$x=\frac{2(pq+1)(p^2-1)}{(p+q{)}^2},y=\frac{2(pq+1)(q^2-1)}{(p+q{)}^2}.$

For integer solutions, set $xy+1=p^2$ and $z=x+y\pm 2p$. Examples:

$p$	$x$	$y$	$z$ values
$3$	$2$	$4$	$12$ or $0$
$4$	$5$	$3$	$16$ or $0$
$5$	$3$	$8$	$21$ or $1$

So integer triples include $(2,4,12)$, $(3,5,16)$, $(3,8,21)$, and $(1,3,8)$.

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Q11 — Generalization with constant $a$ (§232)

Make $xy+a$, $xz+a$, $yz+a$ all squares. Same trick: set $xy+a=p^2$, $z=x+y\pm 2p$ — works for any $a$.

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Q12 (numbered Q11) — Four numbers $x,y,z,v$ with all six pairwise products $+a$ squares (§233-234)

Six formulas $xy+a,xz+a,yz+a,xv+a,yv+a,zv+a$ to make squares. Setting $xy+a=p^2$, $z=x+y+2p$, $v=x+y-2p$ handles five; the sixth $zv+a=(x+y{)}^2-4p^2+a=(x-y{)}^2-3a$ (since $p^2=xy+a$). Setting its root $(x-y)-q$ yields

$x=y+\frac{q^2+3a}{2q},y=\frac{2q{r}^2-2aq}{q^2+3a-4qr},p=y+r$

with $q,r$ free.

A more elegant general approach (§234): represent $x,y,z,v$ via three fractions $b/c,d/e,f/g$ whose pairwise differences have unit numerators, i.e., $be-cd=\pm 1$ etc. Then $x=b^2-a{c}^2$, $y=d^2-a{e}^2$, $z=f^2-a{g}^2$ automatically gives $xy+a=(bd-ace{)}^2$. A table of compatible fractions:

$b/c$	$3/2$	$5/3$	$7/3$	$8/5$	$11/4$	$13/8$	$17/7$
$d/e$	$(3m+1)/(2m+1)$	$(5m+1)/(3m+1)$	$(7m+2)/(3m+1)$	$(8m+3)/(5m+2)$	$(11m+3)/(4m+1)$	$(13m+5)/(8m+3)$	$(17m+5)/(7m+2)$

Combinations of three such fractions handle three numbers; a fourth fraction $h/k=(b+d)/(c+e)$ extends to four.

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Q12 (numbered Q12 in the source, §235) — $x\pm y,x\pm z,y\pm z$ all squares

Six formulas. Suppose $x-y=p^2$, $x-z=q^2$, $y-z=r^2$ — these last three give $q^2=p^2+r^2$, so $(p,r,q)$ a Pythagorean triple: $p=a^2-b^2$, $r=2ab$, $q=a^2+b^2$. Then for the first three formulas, parametrize via biquadrate differences $m^4-n^4=(m^2-n^2)(m^2+n^2)$. A worked example with $f=3,g=9,k=2,h=7$ produces

$x=434657,y=420968,z=150568,$

verifying all six conditions: $x+y=925^2$, $x+z=765^2$, $y+z=756^2$, $x-y=117^2$, $x-z=533^2$, $y-z=520^2$.

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Q14 — Three squares $x^2,y^2,z^2$ with all pairwise differences squares (§236-237)

Want $x^2-y^2$, $x^2-z^2$, $y^2-z^2$ all squares. Divide by $z^2$ and set $x/z=(p^2+1)/(p^2-1)$, $y/z=(q^2+1)/(q^2-1)$. The remaining condition becomes

$\frac{4(p^2{q}^2-1)(q^2-p^2)}{(p^2-1{)}^2(q^2-1{)}^2}=\square .$

After simplification, $(p^2{q}^2-1)(q^2/p^2-1)$ must be a square. Setting $pq=(f^2+g^2)/(2fg)$ and $q/p=(h^2+k^2)/(2hk)$ reduces to making $fg(f^2+g^2)\cdot hk(h^2+k^2)$ a square.

Result: smallest known solution

$x^2=485809,y^2=34225,z^2=23409,$

giving $x^2-y^2=672^2$, $y^2-z^2=104^2$, $x^2-z^2=680^2$.

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Q15 — Three squares with all pairwise sums squares (§238)

Dual problem to Q14. Want $x^2+y^2,x^2+z^2,y^2+z^2$ all squares. Setting $x/z=(p^2-1)/(2p)$ and $y/z=(q^2-1)/(2q)$ resolves the latter two; the first reduces to making $q^2(p^2-1{)}^2+p^2(q^2-1{)}^2$ a square. Several parametrizations give nested solutions; the smallest:

$x=117,y=240,z=44,$

with $x^2+y^2=267^2$, $x^2+z^2=125^2$, $y^2+z^2=244^2$.

A bootstrap remark: any solution $(a,b,c)$ furnishes a new solution $(ab,bc,ac)$, since e.g. $(ab{)}^2+(bc{)}^2=b^2(a^2+c^2)=b^2\cdot \square$.

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Q16 — $x^2+y$ and $y^2+x$ both squares (§239)

Set $x^2+y=(p-x{)}^2=p^2-2px+x^2$, so $y=p^2-2px$; symmetrically $y^2+x=(q-y{)}^2$, so $x=q^2-2qy$. Solving:

$x=\frac{2p^{2}q-q^2}{4pq-1},y=\frac{2p{q}^2-p^2}{4pq-1}.$

Examples: $p=2,q=3$ gives $x=15/23,y=32/23$; $p=1,q=3/2$ gives $x=3/20,y=7/10$.

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Q17 — $x+y$ a square AND $x^2+y^2$ a biquadrate (§240)

Begin with $x^2+y^2$ a square: $x=p^2-q^2$, $y=2pq$. To make this a biquadrate ($r^2+s^2$ raised), set $p=r^2-s^2$, $q=2rs$, so $x^2+y^2=(r^2+s^2{)}^4$. Substitution gives $x=r^4-6r^2{s}^2+s^4$, $y=4r^{3}s-4r{s}^3$.

The remaining condition $x+y=$ square reduces to a quartic in $r,s$:

$r^4+4r^{3}s-6r^2{s}^2-4r{s}^3+s^4=\square .$

Multiple root ansätze produce sample integer solutions. A worked answer:

$x=4,565,486,027,761,y=1,061,652,293,520$

(with $r=1469$, $s=84$, $t=1343$ — Euler emphasizes the size to show the difficulty).

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Concept Pages

• sums-of-two-squares — §217-218 derive that $a+x,a-x$ both squares ⇒ $a$ is a sum of two squares
• ch2.0.13-impossibility-biquadrate-sums — Q8 cites $p^4-q^4\ne \square$ to rule out cases
• brahmagupta-fibonacci-identity — Q5, Q6 are sum-of-two-squares decomposition problems
• infinite-descent — used in §229-230 to prove $p^2+q^2\&p^2+3q^2$ impossibility
• quadratic-residues — §229 uses mod 4, 3, 5 residues
• indeterminate-analysis

Related pages

• ch2.0.4-surd-rationalization
• ch2.0.6-integer-solutions-quadratic-squares
• ch2.0.11-quadratic-form-factorization
• ch2.0.12-quadratic-form-as-power
• ch2.0.13-impossibility-biquadrate-sums
• ch2.0.15-questions-cubes
• pythagorean-triples
• pell-equation